An Introduction to Advanced Mathematics (Short Version)

M. Yotov

April 14, 2020

F or ... Y ou Know W ho!

1 These Notes constitute a version of the course MAA 3200 Introduction to Advanced Mathematics taught by the author at the Department of Mathematics and Statistics of FIU. The concepts of classes, sets, relations, and functions are introduced and studied with rigour. Care is taken in moti- vating the introduction of the Zermelo-Fraenkel axioms. The natural numbers and the principle of (ﬁnite) mathematical induction are discussed in detail. The integer, rational, and real numbers are constructed and thoroughly discussed. As a brief introduction to Advanced Calculus, the classical topology, convergent sequences of real numbers, and continuity of real valued function of a real variable are studied. The Bolzano-Weierstrass Theorem, Intermediate Value Theorem, and Weier- strass’s Theorem are proved.

Please send comments and corrections to the author at yotovm@ﬁu.edu .

c 2017 M.Yotov. Single paper copies for noncommercial personal use may be made without explicit permission from the copyright holder.

2 Contents

1 Logic. Language of Proof 4 1.1 Propositions ...... 4 1.2 Propositional Expressions, Tautologies ...... 6 1.3 Propositional Functions and Quantiﬁers ...... 9 1.4 Methods of Proof ...... 13

2 Sets 21 2.1 Undeﬁned Terms, First Axioms ...... 21 2.2 The Algebra of Sets ...... 27

3 Properties of N 30 3.1 The Least Element Principle (LEP) for N ...... 30 3.2 The Principle of Math Induction ...... 33 3.3 Recursion, + and · in N ...... 36 3.4 What Are Natural Numbers; Peano’s Axioms ...... 38 3.5 Two Examples ...... 40

4 Relations, Functions, and Orders 43 4.1 Relations from a Set to a Set ...... 43 4.2 Functions from a Set to a Set ...... 47 4.3 Equivalence Relations ...... 56 4.4 Orders ...... 60

5 Construction of the Standard Number Systems 67 5.1 The Integers ...... 67 5.2 The Rational Numbers ...... 74 5.3 The Real Numbers ...... 80

6 Topology on the Real Line 89 6.1 The Classical Topology on R ...... 89 6.2 Sequences of Real Numbers ...... 93 6.3 Arithmetic Operations and a Relation on Sequences ...... 100 6.4 Intro to Cantor’s Real Numbers ...... 102

7 The Sets F(X,Y ),X,Y ⊆ R 104 7.1 Limits of Functions ...... 104 7.2 Compact Subsets of R ...... 107 7.3 Compact Subsets of a Topological Space ...... 107 7.4 Continuity of Functions ...... 109

Index 113

3 Chapter 1

Logic. Language of Proof

1.1 Propositions

We begin with an informal discussion of propositions in Logic: the formal approach would lead us too far beyond the scope of the course. In the deﬁnition below, we are relying on the knowledge of the reader about sentences in English. We are also assuming that the reader is able to tell if a statement, i.e., a sentence containing a claim, is true or false. The latter depends on the context in which the statement is considered. In this course, the context will be the one of Mathematics.

Deﬁnition 1.1.1 A proposition P is a statement, i.e., a special sentence, which is either true or false.

It is very important, for our considerations, that there are only two options for a statement: it can either be true or false, and no third option is allowed. Many statements can be formulated (say, in English), but not every such is a proposition (according to the deﬁnition above)! To elevate the status of a sentence from a statement to a proposition, the statement has to be, ﬁrstly, well formulated and understood (all words and symbols in the used in the sentence have to be well understood, and the criteria for telling if the statement if true or false have to be known), and, secondly, the statement has to be either true or false, no third option given. The latter is also expressed by saying that the statement has to have a well determined truth value: T for ”true” and F for ”false”.

Example 1.1.1 1) How are you doing? (This is not a statement) 2) A lunar month is long less than 28 days. (This is a statement, but is not a proposition. The reason is that, for instance, that there are several diﬀerent lunar months, and there is no speciﬁcation of the one in the statement.) 3) The one who is reading these notes now can read English. (This is a statement. It will be a proposition provided we know what ”can read English” means.) 4) There is only one even prime number. (This is a statement. For people who know what even and prime number is, it is a proposition. This proposition has truth value T.) 2

In our course, as already mentioned, we will be considering math propositions only. So, it is important that the sentence, the statement, be formed by using mathematically well deﬁned concepts, and that the claim in that statement be formulated in a mathematically appropriate way. To do that formally, we need to have an alphabet (including math symbols), and also rules how to correctly construct such sentences (so that they be propositions!).

4 Constructing Propositions We explain now how to construct (more complex) propositions starting with given ones P, Q, . . . . To do this we use connectives ∧, ∨, ¬, →, ↔, and delimiters (, ). Thus, one deﬁnes

P ∨ Q, P ∧ Q, ¬P,P → Q, P ↔ Q

by assigning a truth value of each of the constructed propositions based on the truth values of the old propositions. This is conveniently done in truth tables. In words, P ∨ Q (”P or Q”) is false only if both P and Q are false; P ∧ Q (”P and Q”) is true only if both P and Q are true; ¬P (”not P”) is true only if P is false; P → Q (”if P, then Q”) is false only if P is true and Q is false; P ↔ Q (”P if, and only if, Q”) is true only if both P and Q have the same truth value: both are true or both are false.

Example 1.1.2 Consider the statements

P = ”Peter needs help with Calculus 1” Q = ”Peter calls John”

and suppose they are propositions (that is, we know perfectly well what they mean, as well as how to logically evaluate them). To save some further explanations on who Peter and John are, what Calculus 1 is, and what the rules for ﬁnding the truth values of P and Q are, assume P is true, and Q is false. We have the following. • P ∨ Q can be expressed in English as ”Peter needs help with Calculus 1 or he calls John”, or as ”Either Peter needs help with Calculus 1 or he calls John”. This is a true statement. • P ∧ Q can be expressed in English as ”Peter needs help with Calculus 1 and he calls John”. This is a false statement. • ¬P can be expressed in English as ”Peter does not need help with Calculus 1”. This is a false statement. • P → Q can be expressed in English as ”If Peter needs help with Calculus 1, (then) he calls John”, or as ”When/whenever/in case Peter needs help with Calculus 1, he calls John”, or as ”Should Peter need help with Calculus 1, he calls John”, or as ”Peter calls John only if he needs help with Calculus 1”, or as ... This is a false statement. • P ↔ Q can be expressed in English as ”Peter calls John precisely when he needs help with Calculus 1”, or as ”Peter Calls John when he needs help with Calculus 1 and in no other circumstances”, or as ”the only reason for Peter to call John is if he needs help with Calculus 1”, or as ... This is a false statement. The newly deﬁned propositions have names: P ∨ Q is the disjunction of P and Q, P ∧ Q is the conjunction of P and Q, ¬P is the negation/denial of P , P → Q is called a conditional proposition with hypothesis/condition/antecedent the proposition P and with conclusion/consequent - the proposition Q. P ↔ Q is called a bi-conditional proposition.

By using conditionals, one can express every theorem in mathematics. In Logic, the conditional P → Q can be read either as ”Q if P ” (another way of saying ”if P , then Q”), or as ”P only if Q” In English, a conditional can be explained in many, equivalent, but looking quite unrelated, ways. The students have to learn how to detect a conditional in such English expressions.

The conditionals Q → P and ¬Q → ¬P are called converse and contrapositive to P → Q, respectively. The biconditional P ↔ Q is often read as, and replaced by the English expression, ”P if, and only if, Q” or even with the shorter ”P iﬀ Q”.

5 The compound proposition P ∨ Q is called inclusive ”or”. This is, because its value is true not only when P or Q is true, but also when both, P and Q, are true. On the other hand, in everyday speech, when people say ”something” or ”something else”, they usually understand the ”or” exclusively: either ”something” or ”something else”, but not both. Turns out that in math similar to those real life situations appear, and the mathematicians have introduced a special notation for the compound proposition which means that: exclusive or (XOR). Formally, given propositions P and Q, denote by PXORQ the proposition (P ∧ ¬Q) ∨ (Q ∧ ¬P ). The value of this proposition is true if, and only if, exactly one of P and Q is true.

1.2 Propositional Expressions, Tautologies

In our definition of new propositions based on old ones, we didn’t need to know what the old propositions meant. That is, we didn’t need to know what their truth values were. We just exhausted the collection of all combination of values those propositions could have, and determined the value of the new ones. What this formally means is that we simply considered the letters denoting the old propositions as variables who take on values true or false. This suggests the definition of propositional expressions based on an alphabet (p, q, r, ... , x, y, z, ... , X1,X2,X3,...,Y1,Y2,Y3,... ), where we take symbols for our variables-propositions from, connectives (∨, ∧, ¬, →, ↔), and delimiters ((, )), and then using the five ways of constructing compound propositions from the previous subsection. Note that the variables are just symbols. In what follows we will reserve the capital letters from the English alphabet such as P, Q, R, X1,X2,... , to denote propositions known from the context, or variables whose values are propositions. We specifically designate two special symbols, the constants, T and F . to be variables taking only value true in the former case, and only false in the latter case.

Notice the big diﬀerence between (compound) propositions and propositional expressions. The former always have a truth value - they are propositions after all! The latter are not propositions in general! They become propositions after we substitute every variable with a proposition. Asking of the truth value of a propositional expression doesn’t make sense in general. It gets a truth value only after we assign a value to every variable in it. Thus

(”February has 28 days” ∨ ”One lunar month has 28 days”) → ” A week has seven days”

is a proposition, while (P ∨ Q) → R

is a propositional expression. The former has a truth value T (why?), while the question about the truth value of the latter doesn’t make sense: we do not know what P, Q, and R stand for!

So, a bit more formally now, by deﬁnition all symbols from the alphabet, as well as the constants F and T , are propositional expression, and if X and Y are propositional expressions, then so are (X),X ∨ Y,X ∧ Y, ¬X,X → Y , and X ↔ Y .

Note that the propositional expressions are constructed by using only finitely many symbols (variables, constants, connectives, and delimiters). It is reasonable therefore to denote such expressions by P (X1,...,Xn) where X1,...,Xn are symbols from the alphabet - the variables in the expression P . Clearly, every two, and every finitely many, fixed propositional expressions can be considered as depending on the same number of variables. This latter fact allows for a natural logical comparison of propositional expressions.

Deﬁnition 1.2.1 The propositional expressions P (X1,...,Xn) and Q(X1,...,Xn) are called logically equivalent if for every n propositions P1,...,Pn the truth values of P (P1,...,Pn) and

6 Q(P1,...,Pn) are the same. We write in such a case

P (X1,...,Xn) ⇔ Q(X1,...,Xn).

The propositional expressions P (X1,...,Xn) is called a tautology if

P (X1,...,Xn) ⇔ T, and is called a contradiction if P (X1,...,Xn) ⇔ F.

Remark 1.2.1 Note that, by the deﬁnition of the bi-conditional P ↔ Q, we have that the statement

P (X1,...,Xn) ↔ Q(X1,...,Xn) ⇔ T is the same as the statement

P (X1,...,Xn) ⇔ Q(X1,...,Xn). 2

Proving Tautologies/Logical Equivalences; Truth Tables There are basically two ways of proving logical equivalences,

P (X1,...,Xn) ⇔ Q(X1,...,Xn), equivalently - proving tautologies,

P (X1,...,Xn) ↔ Q(X1,...,Xn). One of them is routine, but often a long process. The other is more intelligent, but relies on basic tautologies established using the routine method.

For the ﬁrst one, one compares, for every possible combination (x1, . . . , xn) of values of the variables X1,...,Xn, the values P (x1, . . . , xn) and Q(x1, . . . , xn). The amount of work here depends heavily on both the number of variables, and on the structure of the propositional expressions. There is no much to do when n = 1 for there are two cases to consider: x1 = T and x1 = F . If n = 2, the combinations of values are 4: (T,T ), (T,F ), (F,T ), (F,F ). When n = 3 the combinations of values are 8:

(T,T,T ), (T,T,F ), (T,F,T ), (F,T,T ), (T,F,F ), (F,T,F ), (F,F,T ), (F,F,F ).

The number of possible combinations of values rapidly increases with n being equal to 2n. But even when n is small, like n = 1 or n = 2, it may take long to compute the values of P and Q. For instance, it takes a moment only to realize that X ∧ Y ↔ Y ∧ X is a tautology, while proving that

(X → Y ) ↔ (¬X ∨ Y ) is one needs comparatively more work. To go about the proof that this is indeed a tautology, we use the fact that any such expression is constructed starting with the variables building up using connectives. In our particular case, we gradually construct

¬X,X → Y, ¬X ∨ Y, (X → Y ) ↔ (¬X ∨ Y ).

To compute the value of this expression for a particular choice of values of X and Y , one can compute gradually the values in the building sequence of formulae. Fro instance, if X = T and Y = F , we have for the four propositions above truth values respectively

F,F,F,T.

7 Similar calculations have to be performed for the rest three combinations of values of X and Y .

There is a way to collect all these calculations very neatly in a table, called truth table. • The table has n + k columns where n is the number of variables in the expression we are studying, and k is the number of steps needed to build up this expression. For our example above, n = 2 and k = 4. The first n columns of the table are reserved for the variables involved in the expression. • The rows of the table are 2n + 1: the first row is occupied by the expressions we are evaluat- ing, the rest of the rows correspond to the different combinations of the values of the variables. In our case, the rows are 5. The first n positions of every row contain combinations of values of the variables. The rest of the positions contain the values of the expression placed in the first row at that position. • The table is filled in from left to right, and from top to bottom.

The table corresponding to our example looks as follows

X Y ¬X X → Y ¬X ∨ Y (X → Y ) ↔ (¬X ∨ Y ) T T F T T T T F F F F T F T T T T T F F T T T T

Since the value in the last column are all T, then the expression corresponding to that column, that is, the expression (X → Y ) ↔ (¬X ∨ Y ) is a tautology. Or, formulating it diﬀerently, but equivalently, we have the logical equivalence

X → Y ⇔ ¬X ∨ Y.

We can also put in a truth table the deﬁnitions of the connectives we introduced in this chapter. Here is the table

X Y ¬X X ∧ Y X ∨ Y X → Y X ↔ Y X XOR Y T T F T T T T F T F F F T F F T F T T F T T F T F F T F F T T F

The more intelligent way of proving tautologies/logical equivalences will be demonstrated in the exercises below.

Basic Tautologies The following summarizes the main logical equivalences between propositional expressions.

Proposition 1.2.1 Let P,Q, and R be propositional formulae (of the same number of variables). The propositional formulae are tautologies.

P ↔ PP ∨ P ↔ PP ∧ P ↔ P (¬P ) ∨ P ↔ T (¬P ) ∧ P ↔ F

(P ↔ Q) ↔ (Q ↔ P )(P ↔ Q) ↔ ((P → Q) ∧ (Q → P )) Transitivity of → and ↔

((P ↔ Q) ∧ (Q ↔ R)) → (P ↔ R) ((P → Q) ∧ (Q → R)) → (P → R)

8 Commutativity , associativity, and interaction with the constants T and F

(P ∨ Q) ↔ (Q ∨ P )(P ∨ (Q ∨ R)) ↔ ((P ∨ Q) ∨ R)(P ∨ F ) ↔ P (P ∨ T ) ↔ T

(P ∧ Q) ↔ (Q ∧ P )(P ∧ (Q ∧ R)) ↔ ((P ∧ Q) ∧ R)(P ∧ F ) ↔ F (P ∧ T ) ↔ P Two distributivity laws for ∨ and ∧

(P ∨ (Q ∧ R)) ↔ ((P ∨ Q) ∧ (P ∨ R)) (P ∧ (Q ∨ R)) ↔ ((P ∧ Q) ∨ (P ∧ R))

Negations of the connectives

¬(¬P ) ↔ P ¬(P ∨ Q) ↔ ((¬P ) ∧ (¬Q)) ¬(P ∧ Q) ↔ ((¬P ) ∨ (¬Q))

¬(P → Q) ↔ (P ∧ ¬(Q)) And last, but by no means least!

((¬P ) → (Q ∧ (¬Q))) → P

Proof Exercise (on using truth tables!) 2

Exercise 1.2.1 Prove the claims in Proposition 1.2.1

Remark 1.2.2 It is clear from Proposition 1.2.1 that the connectives ∨ and ∧ behave pretty much like addition and multiplication, respectively, with T being the identity element and F - the zero element. Pay special attention to the last tautology! It can be rephrased as

((¬P ) → F ) → P and, because of that, instrumental for utilizing the method of proof by contradiction to be introduced later in the course. 2

Exercise 1.2.2 (1) Prove that the propositional formula (¬Q ∧ (P → Q)) → ¬P is a tautology. [Proof. We are demonstrating a proof without use of truth tables, but with the use of Proposi- tion 1.2.1 and the tautology we proved before the Proposition 1.2.1. Identify the items from that Proposition we are making use of! We have

(¬Q ∧ (P → Q)) → ¬P ⇔ ¬(¬Q ∧ (¬P ∨ Q)) ∨ ¬P

⇔ (¬(¬Q) ∨ ¬(¬P ∨ Q)) ∨ ¬P ⇔ Q ∨ (¬(¬P ∨ Q) ∨ ¬P ⇔ Q ∨ ¬P ∨ (¬(¬P ∨ Q)) ⇔ (¬P ∨ Q) ∨ ¬(¬P ∨ Q) ⇔ T. Do the exercise using truth tables too, and compare the approaches.]

(2) Prove that the propositional formula ((P ∨ Q) ∧ ¬P ) → Q is a tautology.

(3) Prove that the propositional formula ((P ∨ Q) ∧ ((¬P ) ∨ R)) → (Q ∨ R) is a tautology.

1.3 Propositional Functions and Quantiﬁers 1.3.1 Propositional Functions and Negations thereof Propositional Functions

In forming a propositional expression, say P (X1,...,Xn), the variables X1,...,Xn were allowed to take on values in the collection, the universe, of all propositions. But we can, and will, extend the construction of propositions to include variables taking on values from other universes as well. This

9 will help us construct propositional functions f(X1,...,Xn; x1, . . . , xm) where x1, . . . , xm stand for the variables which are not propositions. Constructing these includes also additional connectives, such as the standard math symbols +, −, ·, ÷, ◦, <, ≤, >, ≥, etc. All these have to be mathematically sound in the most common sense which we assume the reader possesses. A propositional function becomes a proposition if all the variables involved in its construction are given values from their respective universes. A particular case of such functions are the open sentences, which have no propositional variable in them: f(x1, . . . , xm). The non-propositional variables can be of two types: apparent or bound variables, and actual or free variables. Roughly speaking, the apparent variables do not matter in determining the value of the proposition resulting in substituting values for all the variables in f(x1, . . . , xm). For instance, in the propositional function, where the variables x and y are taking on real number values, Z x f(x, y) = ” x cos y dy = 100” 0 the variable x is actual, while the variable y is apparent. As a matter of fact, the variable y, by the very meaning of the formula, is understood to take on values from 0 to x, once x is given a value. So that y is a variable, but we are not allowed to ”assign” values to y. It is in this sense that y is a bound variable. We will see shortly other ways to bound free variables. The function f(x, y) has no propositional variables, so it is an open sentence.

Negations of Propositional Functions

A word about denying a propositional function: ¬f(X1,...,Xn; x1, . . . , xm). The propositional functions are made by using simpler propositional functions and the connectives listed above. We know what to do when denying the logical connectives (this is part of the content of Proposition 1.1). When it comes to the mathematical connectives, such as +, −, ·, ÷, <, ≤, >, ≥, = and others, we notice that • the ﬁrst four of these are binary operations (that is, two elements operated upon by one of these four, produce a new element: given a and b, one gets a + b, a − b, a · b, a ÷ b), and • the last ﬁve are binary relations (that is, they show how two elements are related to each other: a < b, a ≤ b, a > b, a ≥ b, a = b). When a formula is negated/denied, the operations remain the same while the relations get negated So, one has to know how to negate relations. We have in this respect that ¬(<) ⇔ ≥ ¬(≤) ⇔ (>) ¬(=) ⇔ ((<) ∨ (>)) and the three more equivalences obtained by the fact that ¬(¬P ) ⇔ P . As is customary to do, we denote ¬(=) by 6=

Example 1.3.1 Consider the propositions P = ”Year 2025 has 365 days” and Q = ”3 · 15 + 74 ÷ 5 − 15 = π”. Construct a new proposition using P and Q, say R = P → Q. So, R = ”If the year 2025 has 365 days, then 3 · 15 + 74 ÷ 5 − 15 = π.” We know that ¬R = ¬(P → Q) ⇔ P ∧ ¬Q which, in English, sounds like ”Year 2025 has no 365 days, and 3 · 15 + 74 ÷ 5 − 15 is not equal to π”. If we want to formulate the same proposition without using negative expressions, that is, to express it (in positive terms!), then we can write ”Year 2025 has more, or less, than 365 days, and 3 · 15 + 74 ÷ 5 − 15 is bigger, or smaller, than π.” One can use less amount of English words to express the latter as well if they want: ”(year 2025 has more, or less, than 365 days) ∧ ((3 · 15 + 74 ÷ 5 − 15 > π) ∨ (3 · 15 + 74 ÷ 5 − 15 < π)).” Even more variations of expressing this are possible. 2

10 Example 1.3.2 As another example, let’s ﬁnd the negation of the propositional function with integral we considered in the previous subsection. Note that taking an integral of a function is an operation, so a negation should not change it. That is why we have Z x Z x ¬ x cos y dy = 100 ⇔ x cos y dy 6= 100 2 0 0

1.3.2 Quantifiers; Propositions Produced by Quantifying Propositional Functions We know that one way to make a proposition out of a propositional function is by substituting particular values for the variables in the function. In an example above, we figured out that, in the presence of bound variables, it is needed to give values to only the free variables (bounding them this way). In a sense, a propositional function becomes a proposition after all variables in it become bound. A very important other way of achieving the same, making a proposition out of a propositional function, is by using quantifiers. The propositions we get this way are actually claims about the collection of the values of the variables we quantify which make the propositional function have value T . The latter collection is called the truth set of the function. More specifically, consider, as a simplest case, a propositional function of one variable f(x), and let Ωx denote the collection of meanings/witnesses (the universe of discourse) of x. The truth set of f(x) is the collection of all meanings of x for which f(x) is true. The quantification of f(x) produces a claim about the truth set of f(x). Thus, the universal quantification

(∀x)(f(x)) stands for the claim that the truth set of f(x) is the whole Ωx, while the existential quantification (∃x)(f(x)) stands for the claim that the truth set has at least one member. Respectively the symbols ∀ and ∃ are called universal and existential quantifiers. There is also a specialization of the existential quantifier claiming that the truth set of f(x) has only one member in itself (∃!x)(f(x)). This expression is a shorthand for the statement

(∃x)(f(x)) ∧ ((∀y)(∀z)(f(y) ∧ f(z) → y = z)).

A counterexample for (∀x)(f(x)) is a witness x0 for which ¬(f(x0)) is true.

Note that the variable x is a free variable before the quantiﬁcation, and is bound after that. So, the quantiﬁcation is a process of making bound out of free variables.

Having introduced the quantiﬁcations of a function f(x), it is important to know how to interpret the negations of these. The truth of the matter is as follows.

Theorem 1.3.1 For the quantiﬁcations of the propositional function f(x) with universe of discourse Ωx for x we have (1) ¬((∀x)(f(x))) ⇔ (∃x)(¬(f(x)), (2) ¬((∃x)(f(x))) ⇔ (∀x)(¬(f(x)). Proof We are proving the claim in item (1). The claim in (2) is left as an exercise. We have to prove that the propositions ¬((∀x)(f(x))) and (∃x)(¬(f(x)) have the same truth value. For the truth value of ¬((∀x)(f(x))), there are two options: it is either true or it is false. We have to show therefore that if that value is T, then the value of (∃x)(¬(f(x)) is T as well, and that if that truth

11 value is F, then so is the truth value of (∃x)(¬(f(x)). Suppose ﬁrst that ¬((∀x)(f(x))) is true. The proposition (∀x)(f(x)) can either be true or false, Since its negation is true by assumption, then (∀x)(f(x)) is false. This means that not for all x is f(x) true. But this means that there is an x0 ∈ Ωx for which f(x0) is false, that is, there is an x0 ∈ Ωx such that ¬f(X0) is true. The existence of such a x0 can be stated as (∃x)(¬f(x)) is true. Suppose now that ¬((∀x)(f(x))) is false. As we know, this means that (∀x)(f(x)) is true. So, there is no x such that f(x) is false, This means that there is no x for which ¬f(x) is true. In other words, (∃x)(¬f(x)) is false. 2

Exercise 1.3.1 Prove Item (2) of Theorem 1.3.1

Example 1.3.3 Let’s ﬁnd the negation of the proposition (∃ !x)(f(x)). Recall that the formula of this proposition is a shorthand of the (longer) formula

(∃x)(f(x)) ∧ ((∀y)(∀z)(f(y) ∧ f(z) → y = z)).

Therefore

¬((∃ !x)(f(x))) ⇔ ¬[(∃x)(f(x)) ∧ ((∀y)(∀z)(f(y) ∧ f(z) → y = z))]. 2

Exercise 1.3.2 Finish the example above finding the negation of (∃ !x)(f(x)). The case of a function of one (free) variable is naturally extended to the general case of a function of finitely many free variables: f(x1, . . . , xm). In this case, one can, by using quantification, bound any number of variables. So, the result of such quantifications can be a new function (with free variables the unbound yet ones) or a proposition (when all free variables get bound. For instance

(∀x)(∃y)(f(x, y)) is a proposition while (∃x)(f(x, y)) is a function with free variable y. Note that the order of quantiﬁcation does matter! Thus, although (∀x)(∀y)(f(x, y)) ⇔ (∀y)(∀x)(f(x, y)) and (∃x)(∃y)(f(x, y)) ⇔ (∃y)(∃x)(f(x, y)) (prove that as an exercise!),

(∀x)(∃y)(f(x, y)) and (∃y)(∀x)(f(x, y)) are NOT logically equivalent in general!

Example 1.3.4 Suppose Ωx = Ωy = R. The proposition (∀x)(∃y)(x2 = y) is not logically equivalent with (∃y)(∀x)(x2 = y). Indeed, the ﬁrst is a true statement, while the second is false. 2 The theorem above holds true if one out of many free variables only is quantiﬁed. Thus, we have

¬((∀x1)(f(x1, . . . , xn)) ⇔ (∃x1)(¬f(x1, . . . , xn)) and ¬((∃x1)(f(x1, . . . , xn)) ⇔ (∀x1)(¬f(x1, . . . , xn)).

12 Indeed, in this case both sides depend on n − 1 free variables, and the logical equivalence of the o o two means that, for any choice, x2, . . . , xn, of values/witnesses for the free variables, we have that quantiﬁed propositional functions of one variable need to have the same value, namely that

o o o o ¬((∀x1)(f(x1, x2, . . . , xn)) ⇔ (∃x1)(¬f(x1, x2, . . . , xn)) and o o o o ¬((∃x1)(f(x1, x2, . . . , xn)) ⇔ (∀x1)(¬f(x1, x2, . . . , xn)). Since this is true due to the theorem above, then we are done. Now the negation of more general quantiﬁed propositional functions, obtained by quantifying some of the free variables, is formal and straightforward.

Example 1.3.5 (1) ¬((∀x)(∃y)(x2 = y)) ⇔ (∃x)¬((∃y)(x2 = y)) ⇔ (∃x)(∀y)¬(x2 = y) ⇔ (∃x)(∀y)(x2 6= y). This proposition is obviously false (explain why!).

(2) ¬((∃y)(∀x)(x2 = y)) ⇔ (∀y)¬((∀x)(x2 = y)) ⇔ (∀y)(∃x)¬(x2 = y) ⇔ (∀y)(∃x)(x2 6= y). This one is obviously true (explain why!)

(3) Negation of a propositional function with a free variable. ¬((∀x)(x2 = y)) ⇔ (∃x)(x2 6= y).

(4) In this example, we rely on certain knowledge from Calculus I: convergent sequences. Later on in this course, we will study thoroughly this concept. The definition of a convergent sequence is a fine example of a quantified propositional function. We say that the sequence < xn > is convergent if (∃L)(∀)(∃m)(∀n)(n ≥ m → |xn − L| < ).

The propositional function is f(L, , m, n) = ”(n ≥ m → |xn − L| < )”. The universes of meanings for the variables are ΩL = R, Ω = R>0, Ωm = Ωn = N. If the (quantiﬁed) propositional function holds true, then we say that L is the limit of < xn > and write < xn >→ L. A sequence which is not convergent is called divergent. 2

Exercise 1.3.3 Express in positive terms what divergent sequence is.

1.4 Methods of Proof

We are discussing in this section basic methods of proving that a proposition Q is true.

1.4.1 General Remarks A proposition Q in Logic is proved to be true using facts already known to be true. These facts may be axioms (that is, statements that we assume are true without proof), or previously proved propositions. Therefore, proving Q (is true) is formally proving that the conditional

T → Q is true where T stands for an appropriate proposition P known to have truth value T . This leads us to considering methods of proving that a conditional

P → Q is true. Recall that a conditional A → B is false only when A is true, and B is false. So, in our attempt to prove the conditional P → Q is true, it is enough to show that the option when P is true, and Q is false is not possible. Therefore, what we have to actually show is that

if P is true, then Q is true.

13 A proof that a conditional is true may be simple or complicated depending on the complexity of the proposition as well as on our ability to do proofs. Usually, before proving that Q is true, we establish a that several other propositions are true, and these propositions help us ”get” Q. So, we have P → Q1 P ∧ Q1 → Q2 ··· P ∧ Q1 ∧ · · · ∧ Qn → Q. The longer the sequence of conditionals we prove before we prove Q, the more complicated the proof of P → Q. However, there are two speciﬁc schemes of proving a conditional which are used in all proofs: the direct method of proof, and the contrapositive method of proof. The latter one has an important particular case called proof by contradiction. This section is devoted to those three methods of proof. Since all arguments in Logic are constructed using the rules of inference, we begin our discussion by listing, and explaining, some of these.

Vista: Later in this course, in Chapter 3, we will introduce another, quite powerful, method of proof: the Method of (Finite) Mathematical Induction. The three methods discussed in this chapter and the Method of Math Induction constitute the arsenal of methods of proof we will be using in this course. They are the basic ones used in math as well.

1.4.2 Rules of Inference The rules under consideration are just readings of some of the tautologies we have already seen, or interpreting quantiﬁed expressions. Here are some typical examples.

(1) We know that the propositional formula (P ∧ (P → Q)) → Q is a tautology. The rule based on it (modus ponens or law of detachment) says that if P is true, and if P implies Q (that is P → Q is true), then Q is true as well. We are detaching Q from P → Q knowing that P is true. Schematically, this Law can be expressed as

P → Q P ∴ Q (2) Another tautology that we know is the following one: ((¬Q) ∧ (P → Q)) → (¬P ). The rule based on it is called modus tollens (that is, denies by denying). It says that if we know that P implies Q, and that Q is false, then P is false as well. Schematically, we have

P → Q ¬Q ∴ ¬P (3) The hypothetical syllogism rule is based on the transitivity of the conditional:

(P → Q) ∧ (Q → R) → (P → R).

It says that if we know that P implies Q, and that Q implies R, then P implies R. We have P → Q Q → R ∴ P → R (4) The disjunctive syllogism rule is based on the tautology ((P ∨ Q) ∧ (¬P )) → Q , and says that if P is false and if one of P or Q is true, then Q is true.

P ∨ Q ¬P ∴ Q (5) The resolution rule is based on the tautology ((P ∨ Q) ∧ ((¬P ) ∨ R)) → (Q ∨ R). We leave it to the reader (as an exercise) to interpret this rule in English. We write also

14 P ∨ Q ¬P ∨ R ∴ Q ∨ R (6) The universal instantiation rule says that if the proposition (∀x)(P (x)) is true, then for any witness x0 of x the proposition P (x0) is true. We can also write

(∀x)f(x) ∴ f(x0) is true for any x0 (7) The rule in item (6) has a pair, called universal generalization, which says that if for any witness x0 of x the proposition P (x0) is true, then the quantiﬁed proposition (∀x)(P (x)) is also true. So, we have

f(x0) is true for any x0 ∴ (∀x)f(x) The last two rules have their counterparts in the case of the existential quantiﬁcation as well.

Exercise 1.4.1 Formulate the existential instantiation and the existential generalization rules following the model for the universal quantiﬁer.

1.4.3 Direct Method of Proof This is a method for proving that P → Q is true in which one assumes that P is true, and, using laws of inference of Logic and other already proved math facts, proves that Q is true as well. Schematically, the direct method of proof looks like

P ⇒ Q1 ⇒ · · · ⇒ Qn ⇒ Q which is a short way to denote the process of proving that the conditionals in the sequence

P → Q1,P1 → Q2, ··· ,Pn−1 → Qn,Pn → Q are all true. Here P1 = P ∧ Q1,P2 = P1 ∧ Q2,...,Pn = Pn−1 ∧ Qn. Notice that we use the hypothetical syllogism several times to conclude from here that P → Q is true. Notice also that the proof of each of the conditionals in this sequence can be by using a direct method, or any other method of proof we discuss in this course.

Examples of Propositions Proved by Using Direct Method These are given in a form of exercises below. To be able to understand the claims in the examples, we need a bit of deﬁnitions from elementary mathematics: Number Theory and Plane Geometry.

Elementary Number Theory Given two integers, m, n ∈ Z, we say that m divides n, and write m | n, if there is an integer s ∈ Z such that n = ms. We say that n is even if 2 | n, and that n is odd otherwise.

Elementary Plane Geometry According to the axioms of plane geometry (Incidence Geometry) we have that there are points, denoted by capital letters, lines, denoted by small letters, and an incidence relation between them, denoted by ∼ subject to several axioms. Some of them are listed here.

Axiom 1: For every two distinct points, there is a unique line which is incident with both points:

(∀P )(∀Q)(P 6= Q → (∃!l)(P ∼ l ∧ Q ∼ l)).

15 Axiom 2: Every line is incident with at least two distinct points:

(∀l)(∃P )(∃Q)(p 6= Q ∧ P ∼ l ∧ Q ∼ l).

Axiom 3: There are three points such that there is no line incident with all three of them:

(∃P )(∃Q)(∃R)(∀l)(¬(P ∼ l ∧ Q ∼ l ∧ R ∼ l)).

Points incident with a line are called collinear. So, Axiom 3 claims that there are three non- collinear points.

Exercise 1.4.2 (1) Prove that if 0 | n, then n = 0. Prove also that (∀m)(m | 0). [Proof. Note that the ﬁrst statement above, in strict mathematical terms, has to be written as

(∀n)(0 | n → n = 0).

In the proof of both claims, we are using the Universal Generalization rule of inference: we prove the conditional for every (ﬁxed) n, and similarly for the second claim. Starting with the conditional, we have to prove that P → Q has value T where P = ”0 | n” and Q = ”n = 0”. Assuming P is true, and using the deﬁnition of divisibility above, we get that there is an integer number m such that n = 0·m. Now using the fact (which we will prove formally later in the course) that the product of any integer number with 0 is 0, we get that n0 · m = 0. This means that Q is true as well. The claim is proved. Designing a proof of the second claim is left to the reader. ] (2) Consider the claim ”If m | n and n 6= 0, then |m| ≤ |n|”. Formulate it in rigorous mathematical terms, and prove it explaining the rules of inference you are using for that. [Hint: (∀n)(∀m)(m | n ∧ n 6= 0) → (|m| ≤ |n|).]

(3) Consider the claim ”If m | n and m | s, then m | n + s and m | n − s”. Formulate the claim in mathematical terms, and prove it stating the rules of inference you are using to do that. [Hint: (∀m)(∀n)(∀s)((m | n ∧ m | s) → (m | n + s ∧ m | n − s)).]

(4) Prove that (∀m)(∀n)(∀s)((m | n ∧ m | n + s) → (m | s)). State the rule of inference you are using for that. [Hint: Universal Generalization for m, n, s separately. Here P = ”m | n ∧ m | n + s”, and Q = ”m | s”. Assuming P is true, we have that there are natural number u, v such that n = m · u and n + s = m · v. Since n + s = m · u + s, we have m · u + s = m · v. After simple arithmetic manipulations of the last equality, we get s = m · (v − u). By the deﬁnition of divisibility then we get that m | s. The claim is proved. Notice that if (3) above was already proved, the proof of (4) can be very short:

m | n + s ∧ m | n → m | (n + s) − n.]

(5) Prove that if n is even number, then so is n2. Formulate the statement in formal mathematical terms. State the rules of inference you are using in the proof.

(6) Prove that the sum of two even numbers is even. Formulate the claim in formal mathematical terms, and name the rules of inference you use in the proof.

(7)∗ If P is any point, then there is a line which is not incident with P . Formulate this statement in formal mathematical terms. Name the rules of inference you use in the proof.

1.4.4 Contrapositive Method of Proof This is a scheme in which one proves P → Q is true by proving (using direct method) that the conditional ¬Q → ¬P is true instead. This is a valid approach, because as we know the contrapositive

16 and the conditional are logically equivalent P → Q ⇔ ¬Q → ¬P. Often, replacing a conditional with its contrapostive allows for a simpler proof of the theorem at hands.

Examples of Propositions Proved by Using the Contrapositive Method Exercise 1.4.3 (1) Formulate the statement ”if n2 is an odd number, then so is n” in formal mathematical terms, and prove it naming the rules of inference you use for it.

(2) Formulate in strict mathematical terms the claim ”there is at most one point incident with two distinct lines”. Prove the claim stating the rules of inference you use for that. [Proof. We leave the formal statement to the reader. Here P = ”l 6= m”, and Q = ”l and m share no more than one point in common”. We have then that ¬P = ”l = m”, and ¬Q = ”l and m share at least two points in common”. We will show, using direct method, that ¬Q → ¬P is true. To this end, assume ¬Q is true. This means that there are two distinct points, A and B, such that A ∼ l ∧ B ∼ l as well as A ∼ m ∧ B ∼ m. From Axiom 1 of Plane Geometry we get that there is only one line incident with two distinct points. In particular l = m. So ¬P is true. This ﬁnishes the proof.]

1.4.5 Method of Proof by Contradiction (Reductio Ad Absurdum - RAA) This is a very often used method of proof. It is based on the tautology ((¬R) → F ) → R, we encountered and discussed earlier. Using this method amounts to showing that if ¬R is true, this would lead us to ﬁnding a contradiction (proving this way that ¬R → F is true, and therefore, by the tautology above, that R is true).

To show how this method is a particular case of proof by contrapositive, recall that the goal here is to prove R (is true). For this, as we know, one has to prove the conditional T → R (is true). The contrapositive of this conditional is ¬R → F which is exactly what we prove (using direct method) in the proof by contradiction.

Notice that F stands for (some) false proposition, such as S ∧ ¬S, and we actually prove that ¬R → S ∧ ¬S (is true). Schematically, we have ¬R ⇒ · · · ⇒ S ⇒ · · · ⇒ ¬S (⇒ R). There is one thing in this scheme which its users have to be aware of: there are ”appropriate” S, and there are not that ”appropriate” S to work with, and that detecting an appropriate ones depends on the skills and the experience of the person doing the proof! It is even worse, because any proposition can play the role of S. Indeed, if R is true, then its negation ¬R is false, and as we know, no matter what the proposition Q is, the conditional ¬R → Q will be true. So, formally, one will be able to prove that ¬R → S and ¬R → ¬S are true for any S. The art here is to detect S in such a way that it is comparatively easy to prove both ¬R → S and ¬R → ¬S. These are the ”appropriate” propositions S to work with.

17 RAA for Proving P → Q The power of the method of proof by contradiction is most easy to understand when we apply it to proving R = P → Q. The implementation of the tautology ¬R → F looks in this case as follows: one assumes that ¬(P → Q) is true, and proves that, for some proposition S, the contradiction S ∧ (¬S) is true as well. This absurd situation tells us that the assumption was wrong, and that P → Q is true. Note that, as we know already,

¬(P → Q) ⇔ (P ∧ ¬Q).

Comparing this to the direct method of proof of P → Q where one has only one hypothesis (P ), we see that, assuming ¬(P → Q) is true, one gets actually one more hypothesis to work with: in addition to P , the proposition ¬Q is also given to be true. Having these two hypotheses at hand, makes us much more powerful in proving things, so much so that we can even prove S and ¬S for some (and theoretically - for any) S! On the other hand, if the conditional P → Q is actually false, then one will never ﬁnd such an S. If several attempts to ﬁnd an S such that P ∧ ¬Q → S ∧ ¬S is true fail though, it is wise to try to prove that P → Q is actually false.

Examples of Propositions Proved by Using RAA Exercise 1.4.4 (1) If P is any point, then there is a line l which is not incident with P . Formulate this claim in strict mathematical terms, and prove it naming the rules of inference you use for that. [Proof. Let Q = ”(∀P )(∃L)(¬(P ∼ l))”. We want to show that Q is true. This is equivalent, as we know, to proving that Q → T is true. Doing this by RAA means that we assume T ∧ ¬Q, and show that this leads to a contradiction. Since T ∧¬Q = T ∧¬Q ⇔ ¬Q, we see that going by RAA, we assume ¬Q and want to get to a contradiction. As we know, ¬Q ⇔ (existsP )(∀L)(P ∼ l). So, what we assume is that there is a point P such that it is incident with every line. We can easily show that this assumption leads to an absurd situation. Indeed, by Axiom 3 of Plane Geometry we know that there are three points, A, B, C, which are not collinear (do not belong to the same line). Our point P is not equal to at least two of these points (why?). Suppose P 6= A, and denote by l the line determined by A and B, and by m the line determined by A and C.By our assumption, P ∼ l and P ∼ m. So that A ∼ l ∧ P ∼ l and l ∼ m ∧ P ∼ m. By Axiom 1 of Plane Geometry we get that l = m. But then the points A, B and C are all incident with this line! We are done with the proof. Indeed, let S = ”A, B, C are three distinct non-collinear points” where A, B and C are the points provided by Axiom 3. Then, S is a true statement. We proved above that these A, B, and C are also collinear, that is, we proved that ¬S is true as well.]

(2) If l is a line, then there is at least one point P which is not incident with l. Formulate this statement on strict mathematical terms, and prove it naming the rules of inference you use for that.

(3) If P is a point, there are at least two lines incident with it. Formulate this statement on strict mathematical terms, and prove it naming the rules of inference you use for that.

(4) If n is an even number, then n + 1 is an odd number. Formulate this statement on strict mathematical terms, and prove it naming the rules of inference you use for that.

(5) Among two consecutive integers, that is among n and n + 1, one is even, and the other is odd. Formulate this statement on strict mathematical terms, and prove it naming the rules of inference you use for that.

(6) Every odd number has the form n = 2m + 1. Formulate this statement on strict mathematical terms, and prove it naming the rules of inference you use for that.

(7) Prove that if n2 is even, then n is even as well. Formulate this statement on strict mathematical terms, and prove it naming the rules of inference you use for that.

18 (8) The sum of three odd numbers is never 0. Formulate this statement on strict mathematical terms, and prove it naming the rules of inference you use for that. Conclude that if a, b and c are odd integers, then the equation aX2 + bX + c = 0 has no integer solutions. [Proof. We are proving ﬁrst that the sum of any three odd integers is never 0. We do this using direct method (Who are the propositions P and Q here?). Let m, n, and s be odd integers. Then

m = 2m1 + 1, n = 2n1 + 1, s = 2s1 + 1

for some integers m1, n1, and s1. But then

m + n + s = (2m1 + 1) + (2n1 + 1) + (2s1 + 1) = 2(m1 + n1 + s1 + 1) + 1 = 2a + 1

where a = m1 + n1 + s1 + 1. Therefore, m + n + s is an odd number. Since 0 is even, we get that m + n + s is never 0. This completes the proof of the ﬁrst statement. We are proving now the second statement. This can be done using direct method too, but we will will show how to do that by using RAA. (Who are the propositions P and Q here? Who are the two hypotheses we are provided by the RAA?). So, assume that the equation aX2 + bX + c = 0 does have an integer solution where a, b, and c are odd integers. This means that there is an integer n such that an2 + bn + c = 0. Equivalently, we have that c = −an2 − bn. Since c is odd, the integer −an2 − bn is odd as well. Now, there are two options for n: it is either even or it is odd. If n is even, then n2 is even as well (a proven already fact), and both an2 and bn are even. Therefore, −an2 − bn is even. If n is odd, then n2 is odd as well, and both an2 and bn are odd (remember: a, b, and c are odd by the givens!) But then −an2 − bn is even. This ﬁnishes the proof. (Who was the proposition S here?)]

(9)* Suppose a, b, c are integers such that a 6= 0, −1. Prove that one of the polynomials

aX2 + bX + c and (a + 1)X2 + (b + 1)X + (c + 1) has at least one non-integer root. Formulate this statement on strict mathematical terms, and prove it naming the rules of inference you use for that.

Example 1.4.1 Important one. We are proving here a fact which was known to Euclid: there is no rational number whose square is equal to 2. More professionally put, we are proving that the following proposition is true 2 (∀x ∈ Q)(x 6= 2). √ This√ claim is usually formulated as the number 2 is not a rational number. Since the deﬁnition of 2 is a positive real number whose square is 2, we see that the two formulations are equivalent. We will prove the claim by contradiction. To this end, assume

2 (∃x0 ∈ Q)(x0 = 2). We will see later in the course that a rational number can be represented as a reduced rational fraction, that is - as a quotient of two integers, x0 = m/n, where n is a non-zero (actually - a positive) integer, and m and n share no positive factors diﬀerent from 1 We have

m2 m2 = 2 ⇒ = 2 ⇒ m2 = 2n2. n n2 From this it follows that m2 is even, and therefore, as we know from the exercises above, m is even as well: m = 2m1. This in turn gives us

2 2 2 2 2 2 m = 2n ⇒ (2m1) = 2n ⇒ 4m1 = 2n .

19 Simplifying the last equality, by dividing by 2, we get

2 2 2m1 = n

which tells us that n is even as well: n = 2n1. But then m = 2m1 and n = 2n1 share a positive factor diﬀerent from 1 - this contradicts the fact that m/n is a reduced fraction! 2

Remark 1.4.1 The positive integers diﬀerent from one which have only two divisors, 1 and themselves, are called prime numbers. The integer 2 is the smallest prime number. Other such numbers are 3, 5, 7, 11, 13,... . With a bit moire knowledge of the theory of numbers, one can prove that, if √ p is a prime number,√ then p is not a rational number.√ Moreover, it can be proved that, for any positive integer n, if n is not an integer, then n is not a rational number either. 2

1.4.6 Recapitulation on Methods of Proof of P → Q As we saw in this section, every proposition in Logic, and therefore every theorem in Math, can be formulated in the form P → Q. To prove this conditional is true, one needs to establish that if P is true, then Q is true. So, a proof consists of assuming P and proving Q. Making the assumption requires, of course, no much of intellectual power. The whole work goes to do the latter: proving Q (is true). This is done by using known/established already facts and the rules of inference.

In this section, we presented three basic schemes/methods of proving a conditional. Here are, once again, the diﬀerences and the similarities of these.

Suppose we have to prove P → Q (is true). • The Direct Method of proof makes the hypothesis that P is true, and establishes that Q is true. • The Contrapositive Method of proof does a similar thing, but applied to the conditional

¬Q → ¬P.

That is, the hypothesis is ¬Q is true and the conclusion sought is ¬P is true. • The Method by Contradiction does also the same, but applied to the conditional

P ∧ ¬Q → F.

That is, the hypothesis is P ∧ ¬Q is true, and the conclusion sought is S ∧ ¬S is true for an appropriate S. So, in all three methods of proof, one makes a hypothesis, and proves a conclusion employing the rules of inference and already established facts in order to achieve that.

A proof of P → Q, as we have experienced in doing the exercises in this section, consists of proving a chain of (intermediate) conditionals

P → Q1 P1 → Q2 ··· Pn → Q.

Decent proofs, designed to establish deep theorems in math, may consist of hundreds and even of thousands intermediate conditionals! Each of these latter are established using the three methods of proof (and the method of math induction discussed in detail in Chapter 3 of these Notes).

20 Chapter 2

Sets

We are introducing in this chapter, comparatively rigorously, the rudiments of Set Theory. The rest of the course heavily depends on these rudiments: Our point of vie is that every object we deﬁne in Math (such as function, numbers, relations etc.) has to be a set. In our presentation, we will rely on some reasonable intuition as well: the formal introduction of these concepts is a subject of courses on Set Theory taught at FIU.

2.1 Undeﬁned Terms, First Axioms

Sets, denoted by X,Y,... , are intuitively collections of objects having some common features. Notations used to denote such a collection

A = {z | ϕ(z) } where ϕ is a formula deﬁning the property the elements of the collection A have. Mathematically, the situation is a bit more formal.

• We do not deﬁne what sets are. They just are. • There is one relation between sets: the relation ”belong(s) to”, denoted by ∈. We write x ∈ X and read ”(the collection) x is an element of the collection X”.

The slogan here would be that ”there is nothing else, but sets”. That is, the elements of sets are also sets! The negation ¬(x ∈ X) is denoted by x∈ / X.

The situation is actually a bit more complicated: the right slogan is that ”everything is a CLASS”. Collections deﬁned by certain features they possess are not sets in general, they are called classes. Some of the classes are sets. To distinguish between classes and sets in our considerations, we say that sets are classes that exist. We also refer to the classes which are not sets (so, they do not exist) as proper classes.

The ”existence” of some classes is based on the axioms we use to build Set Theory. It is important to emphasize here that the collections we consider are collections of sets. But some collections may be proper classes.

The collection of all sets will be denoted by Ω. The variables in the propositional functions we will be using in the axioms and deﬁnitions below take on values in the collection of sets Ω.

All that said, we are not sure, as of now, if sets exist at all! To make sure they do exist, we need an axiom (the way to go in math). Although the theory can be built by accepting the existence

21 of a set without any further speciﬁcations thereof, we choose to postulate the existence of a special set. In the other approach, it is proved that this special set exists!

Deﬁnition 2.1.1 A set with no elements in it is called empty set.

Axiom 1 (Empty Set Axiom) There is a set with no elements in it.

We denote the set from this axiom with ∅.

Axiom 1 says that empty sets exist. We can not tell at this moment how many these sets are. With the help of some other axioms we will prove shortly that the empty set is unique.

Having assured the sets exist (at least one does!), we would like to know how to work with sets - what new sets we can construct by using old ones. We begin with the deﬁnition of a subset

Deﬁnition 2.1.2 Given two sets X and Y , the set X is a subset of the set Y , we write X ⊆ Y , if (∀x)(x ∈ X → x ∈ Y ).

Exercise 2.1.1 (1) Prove that, for every set X, X ⊆ X. (2) Prove that (∀X)(∀Y )(∀Z)(X ⊆ Y ∧ Y ⊆ Z → X ⊆ Z).

Suppose we have the two inclusions: X ⊆ Y and Y ⊆ X. What can we say about X and Y ? Yes, they have the same elements, and one may be tempted to say that X = Y , that is, that the two sets are identical. But that may not be the case in general! (For example, it may happen that some elements of a set are repeated, and that two sets may have the same elements, but repeated diﬀerent number of times. Such sets need not be considered diﬀerent in general!) We need an axiom to resolve such issues in the theory we are building!

Axiom 2 (Extensionality Axiom) If two sets have the same elements, they are identical . In other words (∀X)(∀Y )((X ⊆ Y ) ∧ (Y ⊆ X) → (X = Y )).

This axiom says, in particular, that the collections we consider are insensitive to having repeated elements in themselves. For instance the sets (we will soon prove these are sets!) {a, b} and {a, a, b}, although having diﬀerent number of elements, are identical for the theory we are developing.

We should be aware though that sets with repeated elements appear naturally, and are widely used, in mathematics. To resolve this apparent issue, one develops the theory of multi-sets. We will brieﬂy discuss this later on in the course in relation with indexed families of sets.

Deﬁnition 2.1.3 The set X is a proper subset of the set Y , denoted by X $ Y , if

(X ⊆ Y ) ∧ ¬(X = Y ).

The negation ¬(X = Y ) is abbreviated, as we know, to X 6= Y .

Here are two theorems that can be proved at this early stage of development of Set Theory. These reveal important properties of the empty set ∅.

Theorem 2.1.1 An empty set is a subset of every set.

22 Proof. Suppose Y is an empty set (a set with no elements). According to the deﬁnition of subset, we have to show that (∀x)((x ∈ Y ) → (x ∈ X)). Since the premise of the conditional in the proposition is false, this proposition is (we say vacuously) true. We are done. 2

Theorem 2.1.2 The empty set is unique: if X is an empty set, then X = ∅. Proof. To prove this theorem we will make a use of Axiom 2 after showing that X ⊆ ∅ ∧ ∅ ⊆ X. According to the deﬁnition of subset, we have to show that

(∀x)((x ∈ X) → (x ∈ ∅)) ∧ (∀x)((x ∈ ∅) → (x ∈ X)).

Both premises of the conditional formulae in the previous line are false: neither ∅ nor X have any elements! Therefore, both conditionals are vacuously true, and the universally quantiﬁed propositions are both true as well. 2

We can not be sure, with the axioms we have so far, that there is a set diﬀerent from the empty set. Here is an axiom which helps us overcome this inconvenience.

Axiom 3 (Power Set Axiom) The collection of all subsets of a set X is a set itself, called the power set of X. In more technical terms

(∀X)(∃Y )(∀z)((z ∈ Y ) ↔ (z ⊆ X)).

Exercise 2.1.2 Use Axiom 2 to prove that the set Y in Axiom 3 is unique. That is if W has the same property as Y from Axiom 3, then Y = W . Notation We denote the set Y in Axiom 3 by P(X).

As a corollary of the Theorem 2.1.1 we get that there are sets diﬀerent from the empty set.

Corollary 2.1.3 (∀X)(P(X) 6= ∅). More precisely, (∀X)(∅ ∈ P(X) ∧ X ∈ P(X)). Proof. By the cited theorem, ∅ ⊆ X, so that ∅ ∈ P(X). This means that P(X) has at lest one element, and therefore is not empty! 2

Having the power sets at our disposal, we can construct plenty of sets: P(∅), P(P(∅)), P(P(P(∅))), etc.

Exercise 2.1.3 (1) List the elements of the sets P(∅), P(P(∅)), P(P(P(∅))). [Hint: P(∅) is a set having only one element - the empty set itself. So P(∅) = {∅}. The set P(P(∅)) consists of the subsets of the set {∅}. There are only two such: ∅, and {∅}. Note that the last two are really sets: the former - by Axiom 1, and the latter - by Axiom 3. So, P(P(∅)) = {∅, {∅}}. Turning to the set P(P(P(∅))), we have to be careful! Here is why. We see that the sets ∅, {∅}, and {∅, {∅}} are subsets of P(P(P(∅))). Note that all the three are indeed sets! Are there more subsets? It is kind of obvious that {{∅}} is also a subset. But why is it a set, to start with? We do not have means to prove that now. For this, we need one more axiom - the Pairing Axiom, to be introduced below. So, ﬁnish the argument here using that Axiom!] (2) Prove that (∀X)(∀Y )(X ⊆ Y → P(X) ⊆ P(Y )), and that (∀X)(∀Y )(X $ Y → P(X) $ P(Y )). The next axioms tell us how to construct even more sets starting from given ones.

Axiom 4 (Pairing Axiom) Given two sets X and Y , there is a set with only elements X and Y . That is, we have (∀X)(∀Y )(∃Z)(∀w)(w ∈ Z ↔ w = X ∨ w = Y ).

23 Exercise 2.1.4 Prove, using Axiom 2, that the set Z the existence of which is declared in Axiom 4 is unique. We denote this set by {X,Y }. In particular, given the set X, there is the set {X,X}. We denote this set by {X}, and consider it as a set of only one element. In general, we call a set with only one element, respectively two (distinct) elements, a singleton, respectively - a doubleton. It follows from Axiom 4 that for every set X, there is a singleton with element X, and that for every two distinct sets X,Y , there is a doubleton with elements X and Y . By using Axiom 2 one easily shows that the singletons and the doubletons with prescribed elements are unique.

In order to be able to construct more sets, we need more axioms. Here is one.

Axiom 5 (Union Axiom) Given a set X, there is a set Y with elements - the elements of the elements of X: Y = {z | (∃x)(x ∈ X ∧ z ∈ x)}.

Exercise 2.1.5 Prove that the set Y above is unique. The set Y is denoted by ∪ X, and is called the union of X. So we have ∪ X = {z | (∃x)(x ∈ X ∧ z ∈ x)}. As an example, ∪{X} = X.

Exercise 2.1.6 Prove that indeed, for any singleton Y = {X}, we have ∪Y = X. As another example, for any two sets X and Y , we have ∪{X,Y } = {z | z ∈ X ∨ z ∈ Y } is a set. We denote the latter by X ∪ Y and call it the union of X and Y .

Exercise 2.1.7 (1) Using Axiom 2, Axiom 4, and Axiom 5 prove that, given sets X,Y and Z, there is a unique set with elements exactly X,Y , and Z. We denote this set by {X,Y,Z}. (2) Compute the power set of the set {X,Y,Z}. Give arguments for your answer. By the way, it is only after this exercise that we are able to prove that the example of sets we used after Axiom 2 is a correct one. Indeed, denote by a = ∅ and b = {∅} (by the way, b = P(∅)). Then the sets {a, b}, and {a, a, b} exist. By Axiom 2 the two sets are equal.

The following axiom is very powerful, and also important. To give an idea why this power and importance, recall that the classes are defined by using a feature their elements have. In notations A = {z | ϕ(z) } where ϕ is a formula encoding the feature. There are two things that we have to be aware of here. The first one is the formula has to be constructed in a special way. The explanation of this goes beyond the scope of these notes, and the reader who wants to learn more about it is referred to any (decent) course on foundations of mathematics. Suffices it to say here that in these notes we use correctly constructed from this point of view formulae. The second thing we have to be aware of, and be careful with, is that even when the formulae are correctly constructed, the collection they define while being a class may or may not be a set. The Axiom 4, and the Axiom 5 for instance assure that when one uses the formulae ϕ(z) = ”z = X ∨ z = Y ” ϕ(z) = ”(∃x)(x ∈ X ∧ z ∈ x)” the corresponding classes are sets.

24 Exercise 2.1.8 Find the formulae used in Axiom 1, and Axiom 3.

Example 2.1.1 Here is a classical example showing that in general the collection defined by a formula is a proper class. This example was discovered by Georg Cantor, the father of Set Theory, popularized later on by Bertrand Russell, and commonly known as the Russell’s paradox. Define the collection M = {z | z∈ / z }. This collection can not be a set! Indeed, if it were a set, then either M ∈ M, and therefore M/∈ M, or M ∈ M, and therefore M/∈ M. (As an exercise, convince yourselves that the previous line is correct mathematically!). So, M is a proper class. It is called Russell’s class. 2 We just proved, by example, that not every class, a collection defined by a formula, is a set. The following axiom helps us avoid situations when the class defined by a formula does not exist.

Here is the promised axiom. It gives a safe way for determining if a class is a set.

Axiom 6 (Separation Axiom) If X is a set, and if P (x) is a propositional function, then the collection {z | (z ∈ X) ∧ (P (z)} is a set. The ﬁrst thing we are doing after introducing Axiom 6 is to prove that intersections of sets exist. We have the following theorem

Theorem 2.1.4 Let X 6= ∅. Then there is a unique set Y such that

(∀z)(z ∈ Y ↔ (∀x)(x ∈ X → z ∈ x)).

We denote this set by ∩X, and call it the intersection of X.

Proof Since X 6= ∅, there is an x0 ∈ X. Then we have

(∀x)(x ∈ X → z ∈ x) ⇔ (z ∈ x0) ∧ (∀x)(x ∈ X → z ∈ x), and then by Axiom 6 the collection Y is a set. The uniqueness of Y is proved by using Axiom 2 in a standard way, and is left to the reader as an exercise. 2

Exercise 2.1.9 Prove that ∪ ∅ = ∅ while ∩ ∅ = Ω. So the latter class doesn’t exist!

Deﬁnition 2.1.4 For any two sets, X,Y , the set ∩{X,Y } is called the intersection of X and Y , and is denoted by X ∩ Y . The sets X and Y are called disjoint if X ∩ Y = ∅. So, we have X ∩ Y = {z | z ∈ X ∧ z ∈ Y }.

The next axiom states that Ω, the collection of all sets, is the same as the Russell’s class M.

Axiom 7 (Russell’s Class Axiom) No set is a member of itself:

(∀x)(x∈ / x).

This axiom allows us to write (keep in mind that the variables in all our propositional formulas are sets!) Ω = {x | x∈ / x}. This exactly means that Ω is ”the same as” the Russell’s class M above. (Note that formally we know how to compare sets only. This is why ”the same as” is in quotation marks. In the end of this chapter we explain how to compare classes as well. Turns out Ω = M.)

25 Remark 2.1.1 (Truth Sets revisited) Now, after we know things about sets, it’s time to revisit our discussion of the truth set of a propositional expression. The truth ”sets” were deﬁned as the collections of all members of the corresponding universes of discourse which satisfy some restriction, e.g., for which the expression was true. But we have to be careful when the universe of discourse is a proper class. For instance, if we deﬁne P (x) = ”x∈ / x”, then we have (b/c of Axiom 7)

{z | P (z)} = Ω which as we now know is not a set! Axiom 6 ensures that when the universe of discourse is a set, then the things are nice: the truth set is a set. 2 By using the above axioms, one can exhibit important constructions of sets. Here is a typical one: the successor of a set.

Deﬁnition 2.1.5 Given a set X, the set S(X) = X ∪ {X} is called the successor set of X.

Exercise 2.1.10 Prove that for every set X, the successor S(X) is really a set. It is the successor which allows us, in particular, to deﬁne the natural numbers! So, as everything that exists in math, the natural numbers are sets. Indeed, we have

∅,S(∅) = {∅},S(S(∅)) = {∅, {∅}},S(S(S(∅) = {∅, {∅}, {∅, {∅}}},... are the sets usually, for short, denoted by 0, 1, 2, 3, etc. In these notations, we have that

0 = ∅, 1 = {0}, 2 = {0, 1}, 3 = {0, 1, 2},...

The class of all natural numbers is denoted by N.

Note that the natural numbers have the following special properties (verify these - Exercise).

• an element of a natural number is either an empty set or is a successor of an element of that number;

• every element of a natural number is also a subset of that number.

Both properties are important in characterizing the natural numbers (as we will see shortly), but the second one turns out to be very useful in developing Set Theory. It has a name:

Deﬁnition 2.1.6 The set X is called transitive if every its element is also a subset of X:

(∀z)( z ∈ X → z ⊆ X).

Exercise 2.1.11 Prove that ∅ is a transitive set. So, the natural numbers deﬁned above are transitive sets.

Exercise 2.1.12 (1) Prove that if X and Y are transitive sets, then so are X ∪ Y and X ∩ Y . (2) Prove more generally that (i) if X is a set with elements transitive sets, then ∪ X is a transitive set as well, and that (ii) if X 6= ∅ is a set with elements transitive sets, then ∩ X is a transitive set. Using Axiom 6, one deﬁnes several sets out of given ones. Here are two examples.

Deﬁnition 2.1.7 Given two sets, X and Y , the relative complement X \ Y of Y in X is deﬁned by X \ Y = {z | z ∈ X ∧ z∈ / Y }.

26 Note: there is no complement of a set in Ω! (Exercise: Prove this!). So, ”absolute” complements to a set do not exist (as sets, of course). But if we agree to call universe some (very big) set, containing all the sets we work with, then this ”absolute” complement exists as well. This follows from Axiom 8 as well.

Definition 2.1.8 Given two sets X and Y , their symmetric difference X ⊕ Y is defined by

X ⊕ Y = {z | z ∈ X ∪ Y ∧ z∈ / X ∩ Y }.

It is important to note at this stage that having in disposal only the Axioms 1 - 7, we are not able to construct sets beyond the realm of finite sets (The proper definition of finite set will be given when we discuss pollency of sets. For now, we use our intuitive understanding of these as sets having finitely many elements.) In particular, we are not able to prove that the class of natural numbers N is a set (in our earlier terminology, we are not able to claim that the class N exists). To make this happen, we need an axiom.

Axiom 8 (Inﬁnity Axiom) There is a set X such that ∅ ∈ X and (∀x)(x ∈ X → S(x) ∈ X). A set satisfying Axiom 8 is called inductive. It can be proved (Do that as an exercise!) that all natural numbers belong to every inductive set. Moreover, the natural numbers in any such set can be distinguished by a formula. Recall that the natural numbers have the following two properties: they are transitive sets, and every element of a natural number is either the empty set or is a successor of a set. So, a formula distinguishing the natural numbers in an inductive set is the following

P (x) = ”(x = ∅) ∨ ((∀z)((z ∈ x) → ((z ⊆ x) ∧ (z = ∅ ∨ (∃y)(y ∈ x ∧ z = S(y)))))”.

By Axiom 6 then, we get that the class N exists: the collection of natural numbers is a set.

Note that the set N satisﬁes Axiom 8. So, N is an inductive set, and, as we saw, it has the distinguishing property of being a subset of every inductive set.

Remark 2.1.2 Often Axiom 8 is stated by simply postulating that the class of natural numbers N is a set. 2

2.2 The Algebra of Sets

In this subsection, we list, and prove as exercises most of them, the main properties of the operations ∩, ∪, \, ⊕. These should be compared to the properties of ∧, ∨, ¬ and XOR respectively.

Theorem 2.2.1 For any given sets X,Y,Z the following holds true.

X ∪ X = XX ∩ X = XX ∪ ∅ = XX ∩ ∅ = ∅ X ∪ Ω = Ω X ∩ Ω = X;

Commutativity and associativity

X ∩ Y = Y ∩ XX ∩ (Y ∩ Z) = (X ∩ Y ) ∩ Z

X ∪ Y = Y ∪ XX ∪ (Y ∪ Z) = (X ∪ Y ) ∪ Z; Distributivity

X ∪ (Y ∩ Z) = (X ∪ Y ) ∩ (X ∪ Z) X ∩ (Y ∪ Z) = (X ∩ Y ) ∪ (X ∩ Z);

De Morgan laws

X \ (Y ∪ Z) = (X \ Y ) ∩ (X \ Z) X \ (Y ∩ Z) = (X \ Y ) ∪ (X \ Z).

27 Proof. Most of these claims are straightforward, and even trivial, to prove. Use deﬁning formulae. 2

Exercise 2.2.1 Prove the identities in Theorem 2.2.1.

Exercise 2.2.2 Prove that (A \ B) \ C = A \ (B ∪ C) = (A \ B) ∩ (A \ C) while A \ (B \ C) = (A \ B) ∪ (A ∩ C) = (A \ B) ∪ (A ∩ B ∩ C). Conclude that the operation relative complement ”\” is not associative in general. [A binary operation ∗ on a set X is called associative if (a ∗ b) ∗ c = a ∗ (b ∗ c).]

Exercise 2.2.3 This exercise, in addition to the theorem above, proves that the operations ⊕ and ∩ have the same properties as addition,+, and multiplication,·, for (integer) numbers as we know them from school. The roles of 0 and 1 are played by ∅ and Ω respectively. Namely, prove that X ⊕ Y = Y ⊕ XX ⊕ (Y ⊕ Z) = (X ⊕ Y ) ⊕ ZX ⊕ X = ∅ X ∩ (Y ⊕ Z) = (X ∩ Y ) ⊕ (X ∩ Z). [For the associativity of ⊕ prove that both sides of the equality are equal to (X \ (Y ∪ Z)) ∪ (Y \ (X ∪ Z)) ∪ (Z \ (X ∪ Y )) ∪ (X ∩ Y ∩ Z). Thus, for the LHS of the equality we have (X ⊕ Y ) ⊕ Z = ((X ⊕ Y ) \ Z) ∪ (Z \ (X ⊕ Y )) = (((X \ Y ) ∪ (Y \ X)) \ Z) ∪ (Z \ ((X ∪ Y ) \ X ∩ Y )) = (((X \ Y ) \ Z) ∪ ((Y \ X) \ Z)) ∪ (Z \ ((X ∪ Y ) \ (X ∩ Y ))) = ((X \ (Y ∪ Z)) ∪ (Y \ (X ∪ Z))) ∪ ((Z \ (X ∪ Y )) ∪ (X ∩ Y ∩ Z)). Do the needed manipulations to the RHS of the equality to get the same expression.]

Remark 2.2.1 The diligent student may have (actually - should have!) noticed that the above interpretation of the class Ω as the identity element for the operations ⊕ and ∩ is not correct: Ω doesn’t exist (it’s a proper class)! Even more than that: having defined intersection (and unioni- sation) for sets only, how should we understand X ∩ Ω (and X ∪ Ω as well)? This is answered the following way. Recall that classes are collections of sets having a common property. This property is given by a propositional function (a formula). For instance, the class Ω is defined by the formula ”x∈ / x”. Then, the union and intersection of classes can be described by formulae as well. Thus, we have X ∪ Ω = {x | x ∈ X ∨ x∈ / x } X ∩ Ω = {x | x ∈ X ∧ x∈ / x }. Now, X being a set, we get that x ∈ X ∨ x∈ / x ⇔ x∈ / x, and therefore X ∪ Ω = {x | x∈ / x } = Ω and we see that a union of a set and a class is not a set in general. Actually such a union is a set if, and only if the class is a set as well (Why?). The case of intersection of a set and a class is much more gentle - by Axiom 6 it always exists! Direct computation shows that X ∩ Ω = X. In any case, if we don’t want to work with proper classes at all, we can consider the family (set!) of all subsets of a fixed set X and apply the operations under discussion to the members of this set only. It is straightforward that all operations on sets we have introduced so far applied to subsets of X produce subsets of X, so that they are well defined in P(X). In this setting, ∅ plays the role of 0, and X - of 1 for the operations ⊕ and ∩. 2

28 2.2.1 A Bit More on Classes Working with classes is not the same as working with sets. Not all constructions possible for sets are possible for classes! We are listing here some of those which extend to the realm of classes.

The classes are deﬁned by a property, given by a formula

A = {z | ϕ(z) }.

Since we consider only collections of sets, z is a set.

We say that the set x is an element of the class A, and write x ∈ A if the proposition ϕ(x) is true.

Suppose A and B are classes, deﬁned by the formulae ϕ and ψ respectively. We say that A is a sub-class of B, and write A ⊆ B, if

(∀x)(x ∈ A → x ∈ B),

that is, if ϕ(x) → ψ(x) is a true proposition. We say that A = B if both A ⊆ B and B ⊆ A. Equivalently, A = B if ϕ ⇔ ψ.

Similarly, union and intersection of classes are deﬁned using disjunction and conjunction of formulae, respectively. In the notations above we thus have

A ∪ B = {x | ϕ(x) ∨ ψ(x)} A ∩ B = {x | ϕ(x) ∧ ψ(x)}. This definition can easily be extended to the case when the classes are finitely many. This can even be done for ”bigger” families of classes, using the corresponding existential and universal quantifications, but we are not going that way in this course. We can also define relative complements, and simmetric difference of classes: just use the corresponding logical operations on the formuale defining the classes.

These deﬁnityions show us that the algebra of classes is actually equivalent/identical to the algebra of logical expressions. But remember that sets are the only classes that exist! These are the classes that matter for math, and this is why the theory of sets is more interesting and harder than the theory of classes (that is, via the deﬁning formulae, more interesting than Logic).

29 Chapter 3

Properties of N

The set N of natural numbers was introduced in the previous chapter using the most important set, ∅, and the operation successor. It was established there that N has interesting properties: • N is an inductive, and a transitive set which is a subset of any inductive set • every non-zero natural number is the collection of all natural numbers constructed before it (by the successor operation).

The first objective of this chapter is to define a relation ≤ on the elements of N, and to show that it satisfies the Least Element Principle (LEP). This important property of (N, ≤) is equivalent, as we prove here, to the Principle of (Finite) Mathematical Induction (P(F)MI) which is the basis of a method of proving some universally quantified propositions. It also allows us to use recursion in constructing sets. As an application of recursion we define two (arithmetic) operations, addition +, and multiplication · on N, and prove that they have properties familiar from the arithmetic classes in elementary school. All this is done in Section 3.1 through Section 3.4.

In Mathematics, and in everyday life, the natural numbers are used for arithmetic purposes. It is not important how they are constructed. Rather, the emphasis is on what we can do using the relation ≤, and the operations + and ·. So, in a sense, the explicitly constructed set N is not what natural numbers really are. We use in Section 3.4 the properties of ≤, +, and · to design a deﬁnition of natural numbers. We show there that the set N is an example of a set of natural numbers, and that any other such set is unambiguously identiﬁable with N.

We close the chapter, Section 3.5, with two statements proved by P(F)MI.

3.1 The Least Element Principle (LEP) for N

We introduce here the relation ≤, and discuss the important property of the relation ≤ on N that it satisﬁes the LEP.

3.1.1 Ordering of the Elements of N The method of ﬁnite induction is based on a property of the natural numbers: the natural numbers can be compared. This leads to introducing a relation on N. The notion of relations between sets will be discussed in much details later in Chapter 4. What we do here will there be considered as an important example of a relation.

Deﬁnition 3.1.1 Let m, n ∈ N. We say that n is less than or equal to m, and write n ≤ m, if (n ∈ m) ∨ (n = m). We say that n is less than m, and write n < m, if n ≤ m ∧ n 6= m.

30 Exercise 3.1.1 Prove that, for m, n ∈ N, we have

n ∈ m ⇔ S(n) ∈ S(m).

[(⇒) Since n ∈ m, there is an m0 ∈ N such that m = S(m0). Therefore, m = {0, . . . , m0}. Note that all elements of m, diﬀerent from m0, have successors in m. Now, there are two options for n ∈ {0, . . . , m0}: either n = m0, and therefore S(n) = S(m0) = m ∈ m ∪ {m} = S(m), or n 6= m0 and therefore S(n) ∈ m ⊂ S(m). In both cases S(n) ∈ S(m). (⇐) S(m) = {0, . . . , m}, and every element of S(m), diﬀerent from 0, is a successor of an element of S(m). Since 0 6= n ∪ {n} = S(n), and having assumed that S(n) ∈ S(m), it follows that n ∈ S(m) being the element whose successor is S(n). We get n ∈ m because n 6= m.]

This exercise shows in particular that S(n) = S(m) only if n = m.

As usual, ¬(n ≤ m) and ¬(n < m) are abbreviated to n 6≤ m and n 6< m respectively.

Note that, since the natural numbers are transitive sets

n ∈ m → m∈ / n.

Exercise 3.1.2 Prove the claim just made!

The following proposition proves that the relation ≤ is a partial order on N In the next chapter, we will discuss relations on sets in general, and partial od=rders in particular.

Proposition 3.1.1 For any three natural numbers m, n, and s, we have (1) n ≤ n; (2) (n ≤ m) ∧ (m ≤ n) → (n = m); (3) (n ≤ m) ∧ (m ≤ s) → (n ≤ s).

Proof (1) Let n ∈ N. Since n = n is true, then n ∈ n ∨ n = n is also true. Therefore, n ≤ n. (2) To prove this item, it is enough to show that if for two natural numbers n and m the proposition (n ≤ m) ∧ (m ≤ n) is true, then n = m. The proposition (n ≤ m) ∧ (m ≤ n) being true means that (n ∈ m ∨ n = m) ∧ (m ∈ n ∨ m = n) is true. This is equivalent to the following disjunction of four propositions being true

(n ∈ m ∧ m ∈ n) ∨ (n ∈ m ∧ m = n) ∨ (n = m ∧ m ∈ n) ∨ (n = m ∧ m = n).

We know already (from the exercise above) that n ∈ m implies that m∈ / n. Therefore, the ﬁrst of the fours propositions is false. The second and the third propositions are false too due to Russell’s Axiom (how?). It follows that the fourth proposition n = m is true as needed. (3) Let now n, m, and s be natural numbers for which (n ≤ m) ∧ (m ≤ s) is true. We have to show that n ≤ s is true as well. Our assumption is equivalent to the following disjunction of four propositions being true

(n ∈ m ∧ m ∈ s) ∨ (n ∈ m ∧ m = s) ∨ (n = m ∧ m ∈ s) ∨ (n = m ∧ m = s).

We consider the four propositions separately, and show that each of them implies the desired n ≤ s. Since natural numbers are transitive sets, we have that m ∈ s implies m ⊆ s. If in addition n ∈ m, then n ∈ s as well. Therefore, (n ∈ m ∧ m ∈ s) implies n ≤ s. Everyone of the second and the third propositions immediately implies that n ∈ s, and so n ≤ s. The last proposition implies n = s, so that again n ≤ s. Item (3) is proved as well. 2

It turns out that every two natural numbers are ≤-comparable. Before proving this, let’s make some observations regarding the natural numbers.

31 Lemma 3.1.2 (1) Let n ∈ N and s, p ∈ n. Then either s = p or s < p, or p < s. (2) Let m, n ∈ N. Then there is p ∈ N such that p = n ∩ m. That is, the intersection of two natural numbers is a natural number itself. Proof (1) The claim is vacuously true if n = 0, and is obvious when n 6= 0 since, in this case, n = S(n0) is a successor of a natural number, and is therefore given by n = {0, 1, . . . , n0}. By the very deﬁnition of the natural numbers, we have that the elements of n are ≤-comparable. (2) The claim is obviously true if n = 0 or m = 0, because in this case n ∩ m = ∅ is a natural number. Let’s assume that n 6= 0 and m 6= 0. By item (1), every two elements of n ∩ m are ≤-comparable. Therefore, there is an element, p0 ∈ n ∩ m which is the biggest there. Since p0 ∈ n and p0 ∈ m, and since natural numbers are transitive sets, we have that p0 ⊆ n and p0 ⊆ m. So, p0 ⊆ n ∩ m. We claim now that n ∩ m = S(p0). Indeed, since p0 ∈ n ∩ m and p0 ⊆ n ∩ m, we have that S(p0) ⊆ n∩m. On the other hand, p0 is the largest in n∩m which implies that every element in the intersection diﬀerent from p0 should be less than it, and therefore should be in p0. We conclude that n ∩ m ⊆ S(p0) as well. This proves our claim. Denoting p = S(p0), we get n ∩ m = p ∈ N. 2

Proposition 3.1.3 The following statement is true (a dichotomy property)

(∀n, m)((n ≤ m) ∨ (m ≤ n)).

Proof Consider the natural number p = n ∩ m. We will prove now that p = n or p = m. The latter is obviously true if n = 0 or m = 0. So, W.L.O.G. we may assume that n 6= 0 and m 6= 0. In this case, p = S(p0) where p0 is the largest element of m ∩ n. Arguing by contradiction now, assume that p 6= n and p 6= m. In this case there are n0 ∈ n \ p and m0 ∈ m \ p. Since every two elements of a natural number are ≤-comparable, and since m0, n0 ∈/ p, we get that p ≤ m0 and p ≤ n0. Now, if p = m0, then m0 ≤ n0 which implies that either m0 = n0 or m0 ∈ n0, in both cases m0 ∈ n ∩ m = p, and so m0 < p which is impossible. Similarly, p 6= n0. Therefore, p < n0 and p < m0 which means that p ∈ m0 ∩ n0. Noticing that n0 ∩ m0 ⊆ n ∩ m, we get that p ∈ n ∩ m = p which contadicts the Russell’s class axiom. This completes the proof that either p = m or p = n. This means that either m ⊆ n or n ⊆ m. This precisely means in turn that either m ≤ n or n ≤ m. 2

The relation ≤ has an important property which, as we will see in the next section, allows the principle of math induction to be a valid method of proof. Namely, we have

Theorem 3.1.4 Every non-empty subset of N has a least element

(∀X)(((X 6= ∅) ∧ (X ⊆ N))) → (∃z)((z ∈ X) ∧ ((∀w)(w ∈ X → z ≤ w))). Proof Since X is a non-empty subset of N, there is a natural number m ∈ X. Consider the set m ∩ X. There are two options for m ∩ X: it is either ∅, or it is distinct from ∅. If m ∩ X = ∅, then all elements smaller that m, being elements of m, are not in X. This exactly means that m is the least element of X. If m ∩ X 6= ∅, then m 6= ∅, so m is a successor: m = S(m0).Therefore m = {0, . . . , m0}, and m ∩ X ⊂ {0, . . . , m0}. Then the smallest element in {0, . . . , m0} which is also in X is the least element of X. 2

The element z ∈ X in Theorem 3.1.4 is called the least element of X. The least element of every non-empty subset of N is unique See the exercise below)! This property of N is called the least element principle or LEP of N.

Exercise 3.1.3 (1) Suppose that m ∈ X ⊆ N. Prove that m is a least element of X if, and only if, m ∩ X = ∅. (2) The least element of a non-empty subset of N is unique. A partial order for which every two elements are comparable, as for ≤, is called a total order.A total order which has the Least Element Property is called a well order. Since N is totally ordered and satisﬁes the LEP with ≤, we say that (N, ≤) is a well ordered set. We will return to well

32 ordered sets in Chapter 4. For now, we will use this property of N to verify a method of math induction (next section).

Exercise 3.1.4 Prove that the relation < has the following properties

(∀n)(n 6< n)(n < m) → (m 6< n) ((n < m) ∧ (m < s)) → (n < s), and (the trichotomy property)

(∀n, m) exactly one of the following holds: n < m, m < n, or n = m.

3.2 The Principle of Math Induction

We describe here a method for proving inﬁnitely many propositions at a time. Recall that the Universal Generalization Rule of Inference teaches that one can prove a statement like (∀x)(P (x)) by proving that for every witness x0 ∈ Ωx the proposition P (x0) is true. The LEP allows us to use another method of proof for such universally quantiﬁed propositional formulae when the universe of discourse of x is a subset of the set of natural numbers:Ωx ⊆ N. Several, equivalent, forms of this method will be given, including reverse induction. In all but one of these a claim about the truth set of a propositional function is stated, and proved. The universes of discourse here are subsets of the natural numbers N. The exceptional version of the Math Induction has to do with the important notion of least elements of subsets of N.

Remark 3.2.1 What we explain here is known as the method of ﬁnite induction. There is a generalization of this, which we will cover if time permits, and which is called method of transﬁnite induction. 2

Immediate Predecessors Recall that for a set X, its successor is deﬁned as S(X) = X ∪ {X}. By the construction of the natural numbers, we see that every non-zero such, n 6= 0, has an immediate predecessor P red(n). More formally, we have

Deﬁnition 3.2.1 Let 0 6= n ∈ N. The immediate predecessor P red(n) of n is deﬁned by m = P red(n) if n = S(m).

Using Exercise 3.1.1, one proves that the predecessor is unique. Here are some properties of P red.

Exercise 3.2.1 (1) Prove that (∀n)(n 6= 0 → (∃ !m)(m = P red(n))). (2) Prove that (∀n)(n 6= 0 → P red(n) = ∪n). (3) Prove that (∀n)(n 6= 0 → n = S(P red(n))). (4) Prove that (∀n)(n = P red(S(n))).

PMI Theorem 3.2.1 The following statements are equivalent1. (1) (∀P (n))(P (0) ∧ (∀n)(P (n) → P (S(n)))) → (∀n)(P (n)), (2) (∀P (n))(P (0) ∧ (∀n)(P (0) ∧ · · · ∧ P (n) → P (S(n)))) → (∀n)(P (n)), (3) (∀P (n))(∀k)((P (k) ∧ (∀n)((k ≤ n) → (P (n) → P (S(n)))) → (∀n)(k ≤ n → P (n)))),

1In the formulation of this theorem we use S(n) instead of the familiar n + 1, because we oﬃcially introduce the operation + for natural numbers only in the next section.

33 (4) (∀P (n))(∀k)((P (k)∧(∀n)((k ≤ n) → ((P (k)∧· · ·∧P (n)) → P (S(n)))) → (∀n)(k ≤ n → P (n)))), (5) (∀P (n))((∀n)(((∀s)(s < n → P (s))) → P (n)) → (∀n)(P (n)), (6) (N, ≤) satisﬁes the LEP.

Remark 3.2.2 This theorem presents six versions of math induction. Although equivalent as statements, the easiness of their applications varies: some of them are more appropriate than others in diﬀerent particular problems at hands. Items (1) through (5) are claims about the truth set of a formula with universe of discourse of its variable - a subset of N. While the claims of items (1) through (4) are traditional, with base cases, the one of (5) has no base cases. This version of math induction is called complete induction. Version (6) is a claim about (N, ≤) - no truth sets here. We know that claim (6) is true. So, claims (1)-(5) are true as well. 2 Proof of Theorem 3.1. We will prove that the conditionals (1) → (2), (2) → (5), (5) → (6), and (6) → (1) are true. This will imply that the propositions of the named four items are all equivalent to each other. As an exercise prove that (1) ⇔ (3) and that (2) ⇔ (4) This will complete the proof of the theorem.

(1) → (2). We are proving here that if (1) is true, then so is (2). Let the propositional formula P (n) satisfy the premise of (2): P (0) ∧ (∀n)(P (0) ∧ · · · ∧ P (n) → P (S(n))). We want to show that the consequent of (2) is true as well. To this end, define the formula Q(n) = P (0)∧· · ·∧P (n). We will show that Q(n) satisfies the premise of (1): Q(0)∧(∀n)(Q(n) → Q(S(n))). Indeed, since Q(0) = P (0), and since P (0) is true, then Q(0) is true. Further, the premise of (2) tells us actually that (∀n)(Q(n) → P (S(n))). Since in general, for any two propositions p and q, we have that p → q ⇔ p → p ∧ q (verify that!), we have that (∀n)(Q(n) → Q(n) ∧ P (S(n))). Since Q(S(n)) = Q(n) ∧ P (S(n)) by the very definition of Q(n), we get that Q(0) ∧ (∀n)(Q(n) → Q(S(n)) that is, Q(n) satisfies the premise of (1) as promised. But then, since we assume (1) is true, the consequent of (1) is also true for Q(n): (∀n)(Q(n)). By the definition of Q(n), we get that (∀n)(P (n)). This proves (2).

(2) → (5). We are assuming (2), and are proving (5) now. So, assume that P (n) satisfies the premise of (5): (∀n)(((∀s)(s < n → P (s))) → P (n)). We will prove that the consequent of (5), (∀n)(P (n)) is also true. Observe now that the proposition (∀s)(s < n → P (s)) is equivalent to T if n = 0, and to P (0) ∧ · · · ∧ P (P red(n))) if n 6= 0. Therefore, the proposition (∀n)(((∀s)(s < n → P (s))) → P (n)) is equivalent to P (0) ∧ (∀n0)(P (0) ∧ · · · ∧ P (n0) → P (S(n0))). This last means that P (n) satisfies the premise of (2). Therefore, it satisfies the conclusion of (2) as well, that is (∀n)(P (n)). This proves (5).

(5) → (6). Assuming (5), we are proving (6) now. We will argue by contradiction. So, assume ∅= 6 Z ⊆ N is such that no element of Z is a least element thereof. Equivalently, Z 6= ∅ ∧ (∀w)(w ∈ Z → w ∩ Z 6= ∅). Therefore, 0 ∈/ Z. Consider the propositional function P (n) = ”n∈ / Z”. We are showing next that the premise of (5), (∀n)((∀s)(s < n → P (s)) → P (n)),

34 is true for this P (n). To this end, using the universal generalization rule of inference, we have to show that, for every natural number n, the conditional

(∀s)(s < n → P (s)) → P (n) is true. But if the hypothesis of the last conditional is true, then no natural number less than n is a member of Z, which is equivalent to n ∩ Z = ∅. But then n∈ / Z according to the hypothesis that Z has no least element. That’s why the consequence of the last conditional, P (n), is also true. Since (5) was assumed to be true, and P (n) satisfies the premise of (5), then P (n) satisfies the consequent of (5) as well: (∀n)(P (n)). This means that no natural number is a member of Z, and so Z = ∅. This contradicts the assumption that Z 6= ∅ which finishes the proof that (6) is true.

(6) → (1). We are proving here, assuming (6), that for any propositional formula P (n) the proposition in (1) is true. Equivalently, for any P (n), the truth set Σ of P (n) satisﬁes Σ = N. We will do this arguing by contradiction. So, let’s assume that there is a formula P (n) which satisﬁes the premise of (1) P (0) ∧ (∀n)(P (n) → P (S(n)), and whose truth set Σ is not the whole set N. Since Σ ⊆ N, the last is equivalent to A := N \ Σ 6= ∅. So we have ∅= 6 A ⊆ N.

We can invoke the LEP from (6) for A:(∃a0)(a0 ∈ A ∧ a0∩A = ∅). Since P (0) is true, we have 0 ∈ Σ, and therefore 0 ∈/ A. This in particular means that a0 6= 0, and therefore a0 = S(P red(a0)). Since a0 is the least element of A, and since P red(a0) < a0, we have that P red(a0) ∈/ A, and therefore that P red(a0) ∈ Σ. The last inclusion is equivalent to P (P red(a0)) being true. By the premise (∀n)(P (n) → P (S(n))) of (1), applied to n = P red(a0), we conclude now that P (S(P red(a0))) = P (a0) is true as well. So, a0 ∈ Σ = N \ A which is impossible since a0 ∈ A. This contradiction proves that Σ = N, and (1) is a true statement for any P (n). 2

Exercise 3.2.2 Prove that (1) ⇔ (3), and that (2) ⇔ (4) in the theorem above.

There is a version of the Principle of Math Induction which is often useful. This version is called reversed or backward induction, and is due to Cauchy.

Exercise-Proposition 3.2.1 Prove that the LEP is equivalent to the following statement

(∀n)(P (2n)) ∧ (∀n)(P (S(n)) → P (n)) → (∀n)(P (n)). 2

Remark 3.2.3 It is obvious that the hypothesis (∀n)(P (2n)) in the proposition above can be replaced by (∀n)(∃m)(m > n ∧ P (m)), or, even better, by ”P (n) is true for inﬁnitely many n ∈ N” (we are still to learn what ”inﬁnitely many” really means.) 2

Vista The transﬁnite induction mentioned above has to do with other (inﬁnite) sets which satisfy the LEP. More precisely, if P (x) is a propositional function with Ωx being a well ordered set with relation ≤, then proving that (∀x)(P (x)) is reduced to proving that

(∀x)((∀y < x)(P (y)) → P (x)), because the following proposition is true

(∀x)((∀y < x)(P (y)) → P (x)) → (∀x)(P (x)).

35 3.3 Recursion, + and · in N Recursion is a method of constructing sets working step by step along well ordered set. We are illustrating this method by deﬁning the operations addition and multiplication on N. We will use recursion later in the course as well - when we discuss statements equivalent to the Axiom of Choice (to be introduced soon).

Deﬁnition 3.3.1 (1) We deﬁne the sum of two natural numbers as follows

(∀m, p ∈ N)((m + 0 = m) ∧ (m + S(p) = S(m + p))). (2) We deﬁne multiplication of two natural numbers as follows

(∀m, p ∈ N)((m · 0 = 0) ∧ (m · S(p) = m · p + m)).

Remark 3.3.1 Obviously, the multiplication is a repeated addition of the same natural number to itself. Also, and this is the mechanism of the recursion, both operations are defined step by step starting with addition (multiplication) of (by) 0, then by 1, 2, etc. That is, the operations are defined based on, already defined, operations with smaller natural numbers. 2

The ﬁrst thing that we have to convince ourselves here is that the addition and the multiplication are deﬁned for every two natural numbers. Indeed, we have the following theorem.

Theorem 3.3.1 For every two natural numbers m and n, the sum m + n and the product m · n are well deﬁned.

Proof Consider the propositional functions

P (n, m) = (∃s)(s = n + m) and Q(n, m) = (∃s)(s = n · m).

Here Ωn = Ωm = Ωs = N. We have to prove that (∀n)(∀m)(P (n, m)) and (∀n)(∀m)(Q(n, m)).

We will do this by using the universal generalization rule of inference applied to n, and then by induction on m. To this end, let n be a natural number, and consider the propositional formulae

Pn(m) = (∃s)(s = n + m) and Qn(m) = (∃s)(s = n · m).

In the new notations, we want to show that the propositions

(∀n)(∀m)(Pn(m)) and (∀n)(∀m)(Qn(m)) are true. As mentioned above, we will work for any (fixed) n, and will prove, by PMI on m, that (∀m)(Pn(m)) and (∀m)(Qn(m)) are true propositions. To do this, we can use any of the equivalent versions of the PMI discussed in Theorem 3.2.1. We are showing that the first proposition, (∀m)(Pn(m)), is true using item (1) of Theorem 3.2.1. The proof of the second proposition, which is very similar to what we do here, is left to the reader as an exercise. Turning to the proof itself, let’s check the base case: m = 0. We have, by the very definition of +, that n + 0 = n, so, choosing s = n, we get that Pn(0) is true. Assume now that Pn(m) is true. We have to prove that Pn(S(m)) is true as well. The induction hypothesis means that n + m is well defined. Denoting by s = S(n+m), and using the definition for addition, S(n+m) = n+S(m), we see that Pn(S(m)) is indeed true. By the PMI we have the proposition (∀m)(Pn(m)) is true. Since this was done for any n, the Universal Generalization Rule of inference gives us that (∀n)((∀m)(Pn(m))) is true. The proof is completed. 2

36 Exercise 3.3.1 Prove the second proposition of the theorem above. Namely, prove that for any natural number n the following is true

(∀m)(∃s)(s = n · m).

Having deﬁned the two operations on N, we have to verify all the properties that we know from school about them. This is done in the exercises below.

Exercise 3.3.2 (1) Prove that the operations addition and multiplication satisfy the expected properties: commutativity, associativity, distributivity of multiplication over addition, 0 and 1 being neutral elements with respect to addition and multiplication, respectively. More precisely, prove that (o) For every natural number m we have

m + 0 = 0 + m m + 1 = 1 + m m · 0 = 0 · m m · 1 = 1 · m

(i) For every three natural numbers m, n, p we have m + (n + p) = (m + n) + p (ii) For every two natural numbers m, n we have m + n = n + m (iii) For every three natural numbers m, n, p we have (n + p) · m = n · m + p · m (iv) For every three natural numbers m, n, p we have m · (n · p) = (m · n) · p (v) For every two natural numbers m, n we have m · n = n · m [Hint: Prove, by diﬀerent forms of induction, the claims in the order they are given above.]

(2) Prove that, for every two natural numbers m, n,

0 6= m ∨ 0 6= n → 0 6= m + n.

The converse of this statement is also true. Prove it.

(3) Provethat both operations, addition and multiplication, have the cancellation property. That is, for every three natural numbers m, n, p we have

m + p = n + p → m = n ((p 6= 0) ∧ (n · p = m · p)) → (n = m).

[Hint: The first claim is proved by contradiction using that a successor natural number is never 0. Prove the cancellation property for addition by induction on p ∈ N. The base case p = 0 is obvious. Assuming that m + p = n + p → m = n is true, let m + S(p) = n + S(p). We want to prove that m = n. But, by the definition of addition, m+S(p) = S(m+p) and n+S(p) = S(n+p). So, we have S(m+p) = S(n+p). By the property of predecessors, it follows then that m+p = n+p, and therefore, by the induction hypothesis, m = n. For the multiplication, using the commutativity of this operation, we will prove, by induction on n, that ((p 6= 0) ∧ (p · n = p · m)) → (n = m). The base case, n = 0 follows by contradiction: if m 6= 0, and p · 0 = p · m, then m = S(m0) and p · m = p · S(m0) = p · m0 + p 6= 0 since p 6= 0. So, 0 = s · 0 = s · m 6= 0 - an absurd! Assuming further that ((p 6= 0) ∧ (p · n = p · m)) → (n = m), consider (p · S(n) = p · m)). We want to prove that S(n) = m. Since S(n) 6= 0, the base case gives that m 6= 0 as well, and m = S(m0). So, we have in this case that p · S(n) = p · S(m0) which, by the definition of multiplication is the following p·n+p = p·m0 +p. By the cancellation property of addition we get that p·n = p·m0 which, by the induction hypothesis gives that n = m0 and therefore that S(n) = S(m0) = m.]

(4) Prove that for the natural numbers m, n, p, the following properties hold true

(0 < n ∧ 0 < m) → (0 < n · m)

(n + p < m + p) → (n < m)(n · p < m · p) → (n < m) (n < m) ⇔ (∃p)(m = n + S(p)).

37 [Hint: The ﬁrst claim follows by contradiction and cancellation (Exercise (3)). For (n + p < m + p) → (n < m) try induction on p. The inductive step is based on the fact that, for a, b ∈ N, if S(a) ∈ S(b), then a ∈ b. For (n · p < m · p) → (n < m) use commutativity of multiplication, and argue by induction on n. Notice that here necesarily we have p 6= 0. For (n < m) ⇔ (∃p)(m = n + S(p)), direction (⇒), do induction on n, using if needed again that S(a) < S(b) → a < b. The direction (⇐) is obvious: n < n + S(p) = S(n + p) = m.]

3.4 What Are Natural Numbers; Peano’s Axioms

We have constructed the set N, consisting of what we called natural numbers, with two operations, addition and multiplication, on it and a relation, ≤. We proved some properties of these as well. The set N has numerous applications in mathematics. None of these applications uses the way natural numbers look like: we just use the symbols like 0, 1, 2, 3.... and some rules how to manipulate with these symbols, e.g. 2 + 3 = 5 or 7 · 5 = 35. The applications of N are actually based on the fact that N is a set which has two operations and a relation with certain properties. So, from this stand point, natural numbers should be the elements of any set having two operations and a relation with the properties like the ones of N.

Speaking of the properties of N, there are some, called defining properties, which determine the rest. Collections of defining properties can be chosen in many ways, depending on the taste of the people who make the choice. We will use one of them, presumably the simplest one, to give a definition of natural numbers. We also include the collection called the Peano’s Axioms - arguably the most popular one.

The discussion below is based on concepts, such as function, injection and bijection which will be introduced later in the Notes. This, we hope, will not make the presentation in this section harder or less understandable though.

3.4.1 A Deﬁnition of Natural Numbers

The basic properties of the set N, which were used to develop the theory of this set, are the existence of the successor operation, S, the element 0 - the only one which is not a successor of an element of N, and the least element principle. The last one is defined using the relation ≤. In order to avoid involvement of a relation as well in the definition of natural numbers, one can replace it with the equivalent, as we know, property of the principle of math induction. Having all this in mind we define what natural numbers are.

Deﬁnition 3.4.1 Natural numbers is a triplet (N˜, 0,S) consisting of a set N˜, and element 0 ∈ N˜ called zero element, and a function S : N˜ → N˜ called a successor function and having the following properties (S1) S is an injection (that is, (∀ x, y ∈ N˜)(x 6= y → S(x) 6= S(y)); (S2) 0 ∈/ Ran(S) (that is, (∀x ∈ N˜)(0 6= S(x))); (S3) (Induction Principle) (∀M ⊆ N˜)((0 ∈ M ∧ S(M) ⊆ M) → M = N˜).

After giving a definition, one has to make sure that the definition is meaningful, that is, that there exists an object, in our case - a set with a zero and a successor function, having the properties (S1), (S2), and (S3). It is straightforward that the set N with the successor operation indeed satisfies these properties. Therefore, the definition of natural numbers is meaningful.

The next thing to be studied is how unique the natural numbers are. There are deﬁnitions which allow only one object to satisfy them. As an example, consider the empty set: we deﬁned it as a set containing no elements (and postulated it exists in Axiom 1). We then proved that this set

38 is unique (we did this by showing that the empty set is a subset of every set, and using the Exten- sionality axiom). In general however, more than one object may satisfy a definition. For example, think of the definition of singleton: a set containing only one element. There are such sets (due to some of the axioms, of course), and there are many singletons: as many as the sets are (since every set X defines a singleton {X}). For another example, there is a well known construction of a set of natural numbers, belonging to Zermelo, in which

0 = ∅, 1 = {∅}, 2 = {{∅}}, 3 = {{{∅}}}, etc.

With this definition, the non-zero natural numbers are all singletons: 0 = ∅, 1 = {0}, 2 = {1}, 3 = {2} etc. When the object satisfying a definition is not unique, the best one can wish for is to have, for any two such objects, an unambiguous way to identify them. Identifications of (distinct) sets are done using bijections between them (these are maps ϕ : X → Y which are injections and surjections, that is, having also that (∀y ∈ Y )(∃x ∈ X)(y = ϕ(x))). The following theorem shows that in the case of natural numbers such identifications exist.

˜ ˜ Theorem 3.4.1 Suppose (N1, 01,S1) and (N2, 02,S2) satisfy the definition of natural numbers. ˜ ˜ Then there is a unique map ϕ : N1 → N2 such that ϕ(01) = 02 and ϕ ◦ S1 = S2 ◦ ϕ. This map is a bijection. ˜ ˜ Proof (Sketch) That one such map exists is straightforward: define f : N1 → N2 recursively as follows • f(01) := 02 ˜ • for every 01 6= x ∈ N1 define f(x) = S2(f(P red1(x))). Using PMI, it is easy to prove that (∀x)(∃y)(y = f(x)), that f ◦S1 = S2 ◦f, and that f is a bijection. ˜ ˜ Here Ωx = N1 and ωy = N2. ˜ ˜ The proof that such a map is unique, one uses induction to prove that, given maps f, g : N1 → N2 satisfying the relations f ◦ S1 = S2 ◦ f and g ◦ S1 = S2 ◦ g and such that f(01) = g(01) = 02, then (∀x)(f(x) = g(x)). The easy proof of this is left as an exercise. 2

˜ ˜ One uses this bijection to identify the triplets (N1, 01,S1) and (N2, 02,S2). The uniqueness of this map makes the identiﬁcation unambiguous, or as we say in such a case - identiﬁes them in a canonical way. So, we may, without a fear of ambiguity, denote (any) set of natural numbers by the familiar N. The same applies to the operations and the relation of a set of natural numbers.

3.4.2 Peano’s Axioms An equivalent characterization of the natural numbers is given by using the Peano’s Axioms. Namely we have

Theorem 3.4.2 Suppose the set X with the properties (P1) (∃x)(x ∈ X) (P2) (∀y)(∃y0 ∈ X) (P3) ¬((∃y ∈ X)(y0 = x)) (P4) (∀y, z ∈ X)(y0 = z0 → y = z) (P5) (∀Y ⊆ X)((x ∈ Y ∧ (∀w ∈ X)(w ∈ Y → w0 ∈ Y ) → Y = X). Then the triplet (X, x, (·)0) satisfies the properties of the definition of natural numbers. Conversely, if the triplet (N˜, 0,S) satisfies the definition of natural numbers, then the set X = N˜ has the properties (P1)-(P5) with x = 0 and y0 = S(y).

Proof The easy, but long proof is left as an exercise to the diligent readers. 2

The propositions (P1)-(P5) are called Peano’s axioms. The set of natural numbers N we have constructed does satisfy the Peano’s axioms (as we have proved along the way).

39 3.5 Two Examples

This section is devoted to two interesting examples of theorems proved by PMI. The ﬁrst theorem is a statement about integers, and the second is a statement about real numbers. As we know, every object in math has to exist as a set. The integers and the real numbers are no exception. In Chapter 5, using the set N, and some set theoretical constructions, we will deﬁne the basic number systems (integers, rational numbers, real numbers), and will prove that they exist. For the needs of this section though we will assume knowledge of the basic properties of the integers and the real numbers, so that we can state and prove the promised two theorems.

Long Division for Integers

The ﬁrst theorem we are proving is the following one.

Theorem 3.5.1 (Long Division for Integers) Suppose n, m ∈ Z where m 6= 0. Then there are unique q, r ∈ Z such that n = m · q + r, 0 ≤ r < |m|.

Proof. (Existence of q and r) Consider the set Σ = {n − m · q | q ∈ Z}. Since m 6= 0, the set Σ contains non-negative integers. Indeed, if n ≥ 0, then n − m · 0 is a non-negative integer in Σ. If n < 0, and m > 0, then n − m · n is a positive integer in Σ. If, ﬁnally, n < 0 and m < 0, then n − m · (−n) is a positive integer in Σ. This implies that the set S = Σ ∩ N is a non-empty subset of N, and therefore has a least element r ∈ S. The latter means that there is an integer q such that r = n − m · q. In other words, n = m · q + r. We claim now that r < |m|. Indeed, if not, that is if r ≥ |m|, then 0 ≤ r0 = r − m · q − |m| < r and r0 ∈ S which is impossible due to r being the least element of S. So, we found integers q and r such that n = m · q + r and 0 ≤ r < |m|. (Uniqueness of q and r) Assume that the integers q0 and r0 have the properties that n = m · q0 + r0 and 0 ≤ r0 < |m|. we have to show that q0 = q and r0 = r. We have

m · q + r = m · q0 + r0(= n).

Therefore m · (q − q0) = r0 − r

which implies that |m| · |q − q0| = |r0 − r|.

Assume now, by way of RAA, that either q 6= q0 or r 6= r0. Since m 6= 0, and using the last equality, we see that we have both r 6= r0 and q 6= q0. but then |q − q0| ≥ 1, and

|r0 − r| = |m| · |q − q0| ≥ |m| · 1 = |m|.

On the other hand, the inequalities 0 ≤ r < |m| and 0 ≤ r0 < |m| imply that 0 ≤ |r0 − r| < |m| (do you see why?), and we get |m| ≤ |r0 − r| < |m|

and ultimately that |m| < |m| which is a contradiction! Therefore, by RAA, we get that q = q0 and r = r0. The theorem is proved. 2

Deﬁnition 3.5.1 The number q is called the quotient, and the number r is called the remainder of the division of n by m. The number n is called the dividend, and the number m is called the divisor of that division. Finding q and r is called long division process.

40 A Classical Inequality For the next result, we assume knowledge of real numbers, and of taking nth roots of (non-negative) real numbers.

Theorem 3.5.2 (AM-GM Inequality) For any non-negative real numbers a1, a2, . . . , an, the following result holds true a1 + a2 + ··· an √ ≥ n a · a ··· a , n 1 2 n

and the equality happens if, and only if, a1 = a2 = ··· = an. Proof. (Sketch) The proof is a classical example (due to Cauchy) of application of the reverse induction. One ﬁrst proves, using a simple induction, that the inequality is true for every n = 2k where k ≥ 1. The base case here, k = 1 is an easy thing to verify (do that!), while the inductive step goes as follows

a + ··· + a k+1 1 (a + ··· + a k ) + (a k + ··· + a k+1 ) 1 2 = · 1 2 2 +1 2 2k+1 2 2k 1 a + ··· + a k a k + ··· + a k+1 1 √ √ 1 2 2 +1 2 2k k · + ≥ · a ··· a k + 2 a k ··· a k+1 2 2k 2k 2 1 2 2 +1 2

q √k √ k+1√ 2 2k 2 ≥ a1 ··· a2k · a2k+1 ··· a2k+1 = a1 ··· a2k+1 . Having proven this, one shows next, and this is the reverse induction step, that if the inequality is true for n + 1 ≥ 2 , then it is true for n as well. This is achieved by applying the inequality to the a1+···+an numbers a1, . . . , an, n noticing that

a + ··· a a + ··· + a + a1+···+an 1 n = 1 n n n n + 1 and that √ q √ n n+1 n a1 ··· an = a1 ··· an · a1 ··· an .

Finish the proof. Don’t forget to show that the equality happens if, and only if, a1 = ··· = an. 2

Deﬁnition 3.5.2 The claim in the last theorem is called the arithmetic-geometric mean inequality, and is abbreviated to AM-GM.

Exercise 3.5.1 (1) Prove that for every natural n

n(n + 1) 1 + 2 + ··· + n = . 2 (2) Let a and b be natural numbers. Prove that, for every natural number n,

(2a + nb)(n + 1) a + (a + b) + ··· + (a + nb) = . 2 (3) Let x be a real number such that x 6= 1. Prove that, for every natural number n,

1 − xn+1 1 + x + x2 + ··· xn = . 1 − x (4) Let a and x be real numbers such that x 6= 1. Prove that, for every natural number n,

1 − xn+1 a + ax + ax2 + ··· + axn = a . 1 − x

41 (5) Prove that, for every natural number n,

n(n + 1)(2n + 1) 12 + 22 + ··· + n2 = . 6 (6) Prove that, for every natural number n,

n(n + 1)2 13 + 23 + ··· + n3 = . 2

(7) (Bernoulli’s inequality) Let x be a real number such that x ≥ −1. Prove that, for every natural number n, (1 + x)n ≥ 1 + nx. (8) Prove that, for every odd natural number n ≥ 3,

(1 + 1/2)(1 − 1/3) ··· (1 + (−1)n/n) = 1, and that for every positive even natural number n,

(1 − 1/2)(1 + 1/3) ··· (1 − (−1)n/n) = 1/2.

(9) Find a formula for the product below, and prove (by induction or otherwise) it is true

(1 − 1/4)(1 − 1/9)(1 − 1/16) ··· (1 − 1/n2).

(10) The same as in item (9), but for the expression

1 · 1! + 2 · 2! + ··· + n · n!.

42 Chapter 4

Relations, Functions, and Orders

After having studied sets statically, as separate entities, in the previous two chapters, it is time to make the picture more dinamic, and introduce the means to compare sets. To this end, we are introducing and studying in considerable detail the concept of a relation from a set to a set gradually specializing it to multi-valued functions, partial functions and finallly - to functions from a set to a set. All these are, of course, sets, and basic set operations (union, intersection), as well as certain specific constructions (restriction, inversion, composition) can be applied to them. In all four cases, we find the conditions under which the listed operations applied to objects of certain type do not lead us out of that type. This, in the case of the operation union, leads us to conditions under which (partial) functions can be glued, and in the case of inversion naturally leads us to the concept of one-to-one patrtial functions, and to one-to-one correspondences. We pay special attention to the most important types of functions: injections, surjections, and bijections, and study them through their compositions with other functions. This way we get to the concept of left and right inverses of a function characterizing injections and surjections. Our attempt to characterize the surjections in this context leads us to introducing the Axiom of Choice.

Substantial part of the section is devoted to relations on a set, that is, to relations from a set to itself. While relations from a set to another one gives the means to compare these sets, relations on a set provide it with important structures. In this course we are interested mainly in equivalence relations and diﬀerent types of orders on a set. Using the former we will construct in Chapter 5 sets of integer and rational numbers, while using the latter we will construct there a set of real numbers.

4.1 Relations from a Set to a Set

4.1.1 Ordered Pairs of Sets The basic concept here is: ordered pair of sets. The following deﬁnition is due to the Poland Set Theorist and Topologist Kazimierz Kuratowski.

Deﬁnition 4.1.1 Let x, y be sets. The set

(x, y) := {{x}, {x, y}} is called the ordered pair with ﬁrst component x and second component y.

Exercise 4.1.1 Specify the axioms we need to use in order to deﬁne the ordered pair (x, y).

43 The most important property of the ordered pairs is that the two components are distinguishable set theoretically (not only graphically). We have the following fact.

Proposition 4.1.1 For any four sets x, y, x1, y1,

(x, y) = (x1, y1) ⇔ (x = x1) ∧ (y = y1).

Proof (⇒) Suppose (x, y) = (x1, y1). Note ﬁrst that (x, y) is a singleton if, and only if, x = y, in which case (x, y) = {{x}}. Indeed, {{x}, {x, y}} is a singleton only if {x} = {x, y}, only if x = y. In such a case, {{x}, {x, y}} = {{x}, {x, x}} = {{x}, {x}} = {{x}}. We consider next two cases. (1) x = y (that is, (x, y) = {{x}} is a singleton). Then (x1, y1) is a singleton as well, and therefore x1 = y1 with (x1, y1) = {{x1}}. The relation (x, y) = (x1, y1) means that {{x}} = {{x1}} which is possible only when {x} = {x1}, or when x = x1. We also have x = x1 and y = x = x1 = y1, so y = y1. (2) x 6= y (that is, (x, y) is a doubleton). In this case (x1, y2) is a doubleton as well, and hence x1 6= y1. Now the equality of sets

{{x}, {x, y}} = {{x1}, {x1, y1}} is possible only if either {x} = {x1} and {x, y} = {x1, y1}, in which case we are done, because x = x1 and (since x 6= y and x1 6= y1) y = y1, or

{x} = {x1, y1} and {x, y} = {x1}.

But this latter case is impossible, because the ﬁrst equality implies that x1 = y1(= x) which is not the case. (⇐) This direction of the claim is obvious. 2

Remark 4.1.1 There are several definitions of ordered pair. All they are designed to distinguish between two elements making the pair. The most commonly used definition different from ours (see for instance the book on Algebra by P. Aluffi cited in the end of these Notes) is this (we decorate the notation to distinguish it from the one above) (x, y)0 = {x, {x, y}}. As in the case of Kuratowski’s definition, it is true that 0 0 (x, y) = (x1, y1) ⇔ x = x1 ∧ y = y1, but the proof is more involved that the one for (x, y), and uses an axiom we haven’t introduced here (the axiom and a sketch of a proof of the fact above is given in the last section of this chapter). This is one of the reasons we used Kuratowski’s definition of ordered pair. 2

Exercise 4.1.2 Prove that for no x and y is (x, y) = (x, y)0. Repeating the construction of ordered pair, one deﬁnes ordered triplets, quadruplets, etc.

Exercise 4.1.3 (1) Deﬁne ordered triplets of sets by (x, y, z) = ((x, y), z). Prove that

(x, y, z) = (x1, y1, z1) ⇔ x = x1 ∧ y = y1 ∧ z = z1. (2) Similarly, deﬁne (x, y, z)00 = (x, (y, z)). Prove that 00 00 (x, y, z) = (x1, y1, z1) ⇔ x = x1 ∧ y = y1 ∧ z = z1.

Remark 4.1.2 It is a natural question to ask when is (x, y, z) = (x, y, z)00? In other words, when is ((x, y), z) = (x, (y, z))? With the introduced so far axioms we are not able to answer this question. More precisely, having the equality of the two sets is consistent with our axioms. In the end of this chapter, where we discuss a stronger version of the Russell’s class axiom, we will show that with this axiom replacing the Russell’s class axiom we can prove that it is actually never true that (x, y, z) = (x, y, z)00. 2

44 4.1.2 The Cartesian Product of Two Sets

Proposition 4.1.2 The following holds true

(∀A)(∀B)(∃!C)((∀z)(z ∈ C ⇔ (∃a)(∃b)(z = (a, b)), where Ωa = A, Ωb = B, and as usual A and B are sets.

Proof The existence of C follows from the observation that {a}, {a, b} ∈ P(A ∪ B), and therefore that (a, b) ∈ P(P(A ∪ B)). Indeed, by the Separation Axiom (Axiom 6), the class C is deﬁned by a formula from the set P(P(A ∪ B))

C = {z | z ∈ P(P(A ∪ B)) ∧ (∃a)(∃b)(a ∈ A ∧ b ∈ B ∧ z = (a, b))}.

The uniqueness of C is derived in a standard way by using the Extensionality Axiom (Axiom 2). 2

The set C is so important (as we will soon see), that it has a name.

Deﬁnition 4.1.2 The Cartesian product, X × Y , of two sets X and Y is deﬁned by

X × Y = {(x, y) | x ∈ X ∧ y ∈ Y }.

Exercise 4.1.4 (1) Prove that, in general, X × Y 6= Y × X, but that, for any A,

A × ∅ = ∅ × A = ∅.

(2) Prove more generally that if X 6= ∅= 6 Y , then

X × Y = Y × X ⇔ X = Y.

(3) Prove that the collection

{(x, y, z) | x ∈ X ∧ y ∈ Y ∧ z ∈ Z } is a set, and is uniquely determined by X,Y , and Z. We denote this set by X × Y × Z. Prove more speciﬁcally that X × Y × Z = (X × Y ) × Z.

4.1.3 Relations from a Set to a Set The following deﬁnition is fundamental for our considerations.

Deﬁnition 4.1.3 Given two sets A and B, a relation from A to B is a triplet (A, B, R) such that R ⊆ A × B. A relation from A to A is called a relation on A.

Note that, since R ⊆ A × B is a collection of ordered pairs, it is a relation on its own. As such, it has domain and range:

Dom(R) = {a | (∃b)((a, b) ∈ R)} Ran(R) = {b | (∃a)((a, b) ∈ R}.

Therefore, Dom(R) ⊆ A and Ran(R) ⊆ B. Note that both inclusions may be strict.

In the case of relations from A to B, we have the following terminology.

Deﬁnition 4.1.4 Given (A, B, R), the set B is called the co-domain of (A, B, R). We write co − Dom(R) = B.

45 Deﬁnition 4.1.5 Given a relation (A, B, R) and x ∈ A, deﬁne

R[x] = {y ∈ B | (x, y) ∈ R }.

Obviously, R[x] ⊆ B are well deﬁned for every set x. Of course, R[x] is non-empty only if x ∈ Dom(R).

Deﬁnition 4.1.6 The relation (B, A, S) from B to A is called the inverse relation of (A, B, R), if S = {(b, a) | (a, b) ∈ R }. We may also write (B, A, R−1) = (B, A, S) = (A, B, R)−1.

From now on in these notes we will be abbreviating the notation (A, B, R) to R. Keep in mind that two relations (A, B, R) and (C,D,S) are equal if, and only if, A = C,B = D, and R = S. So, using the abbreviation we have to be aware of who the sets involved are. We will be ﬁxing this by writing R ⊆ A × B.

Example 4.1.1 (1) For every two sets, there always is relation from A to B: this is ∅ ⊆ A × B. If A × B 6= ∅, there always is a second relation: A × B ⊆ A × B. (2) A very important relation on a set A is the identity relation iA ⊆ A × A

iA := {(x, x) | x ∈ A }. 2

Exercise 4.1.5 (1) Prove that, for any relation R ⊆ A × B, (R−1)−1 = R. (2) Prove that R[x] exists (i.e., is a set) for every set x. (3) Let R ⊆ A × B. Consider the set {R[x] | x ∈ A}. Recall that in it, we do not distinguish between R[x] and R[x0] if (x, x0) ∈ R. Similarly, consider {R−1[y] | y ∈ B}. Prove that

Ran(R) = ∪{R[x] | x ∈ A} Dom(R) = ∪{R−1[y] | y ∈ B}.

Restriction, Intersection, Union, and Composition of Relations from a Set to a Set The restriction of relations from a set to a set is quite a ﬂexible operation. It is not always a restriction in the intuitive sense we have from school (or life).

Let R ⊆ A × B be a relation, and let C and D be two sets.

Deﬁnition 4.1.7 The restriction of R to C × D, denoted by R|C×D, is deﬁned by

R|C×D := R ∩ (C × D) ⊆ C × D.

If R is a relation on A, the restriction of R to C, denoted R|C , is deﬁned by

R|C = R|C×C ⊆ C × C. Note that the restriction of a relation is a way to deﬁne a relation between two new sets, the relationship with the original ones being not important. Thus, if B ⊂ D then the restriction expands the co-domain of the original relation. The options here are many. Also, keep in mind that the restricted relation is equal to the original one only if A = C and B = D. (B/c this is the case if (A, B, R) = (C,D,R|C×D).)

Exercise 4.1.6 (1) Prove that (R|C×D)|E×F is the relation R ∩ ((C ∩ E) × (D ∩ F )) ⊆ E × F.

Conclude that, in general, (R|C×D)|E×F 6= (R|E×F )|C×D. (2) Suppose C ∩ Dom(R) = ∅ or D ∩ Ran(R) = ∅. Compute the restriction R|C×D.

46 The operations intersection and union of sets provides operations on relations from a set to a set as follows.

Deﬁnition 4.1.8 Let (A, B, R) and (C,D,S) be two relations. (1) The intersection of these relations is deﬁned by

(A, B, R) ∩ (C,D,S) := (A ∩ C,B ∩ D,R ∩ S).

(2) The union of the relations is deﬁned by

(A, B, R) ∪ (C,D,S) = (A ∪ C,B ∪ D,R ∪ S).

An operation typical for relations (not available for sets which are not collection of ordered pairs) is the composition of relations.

Deﬁnition 4.1.9 Let R ⊆ A × B and S ⊆ C × D.The composition, S ◦ R ⊆ A × D, of R and S is deﬁned by S ◦ R := {(x, z) | (∃y)((x, y) ∈ R ∧ (y, z) ∈ S } ⊆ A × D.

Exercise 4.1.7 (1) Prove that, for every R ⊆ A × B,

R ◦ iA = iB ◦ R = R;

(2) Suppose L is a relation on A such that, for every R ⊆ A × B and for every S ⊆ B × A,

R ◦ L = R and L ◦ S = S.

Prove that L = iA. (This shows that the relation iA can be characterized by using composition of relations only!) (3) Prove that the composition of relations is associative, that is, for any R ⊆ A × B,S ⊆ B × C, and T ⊆ C × D, T ◦ (S ◦ R) = (T ◦ S) ◦ R.

Remark 4.1.3 We see that the ﬁve operations, inversion, restriction, intersection, union, and composition, are available in the case of relations from a set to a set producing relations from a set to a set. 2

4.2 Functions from a Set to a Set

Most of the relations from A to B, that is, most of the elements of P(A × B), are not interesting as providing information about the interaction of the two sets. In this subsection, we restrict our attention to relations obeying some restrictions. For R ⊆ A × B, these restrictions are described in terms of Dom(R), Ran(R), and R[x] for x ∈ A.

4.2.1 Multivalued Functions and Correspondences Deﬁnition 4.2.1 The relation (A, B, R) is a multi-valued function if Dom(R) = A. A multi- valued function is called a correspondence from A to B if in addition Ran(R) = B. Obviously, R ⊆ A × B is a multi-valued function if, and only if, (∀x)(x ∈ A → R[x] 6= ∅), and is a correspondence, if, and only if, the inverse R−1 ⊆ B × A is a multi-valued function as well. To denote multi-valued functions, we some times write

R : A ⇒ B instead of the usual ”R ⊆ A × B is a multi-valued function”.

47 Example 4.2.1 (1) The relation (A, A, iA) is a multivalued function (2) The relation (∅, A, ∅) is a multivalued function, but (A, ∅, ∅) is not if A 6= ∅. (3) The relation (A, B, A × B) is a multivalued function if B 6= ∅. (4) The relation ({x},B,R) is a multivalued function if, and only if, R 6= ∅. 2

Exercise 4.2.1 Prove the following (1) The union of two multi-valued functions is a multi-valued function. (2) Let R : A ⇒ B and S : C ⇒ D be such that (∗)(∀a ∈ A)(R[a] ∩ C 6= ∅). Prove that the composition S ◦ R ⊆ A × D is a multi-valued function as well. Prove also that the condition (∗) is not only suﬃcient, but necessary as well in order to have the composition a multi- valued function. (3) Let R : A ⇒ B and S : C ⇒ D be such that (∗∗)(∀z ∈ A ∩ C)(R[z] ∩ S[z] 6= ∅). Prove that the intersection (A ∩ C,B ∩ D,R ∩ S) is a multi-valued function as well. Prove also that the condition (∗∗) is not only suﬃcient, but necessary as well in order to have the intersection a multi-valued function. −1 (4) Consider R : A ⇒ B. Prove that if R is a correspondence from A to B, then iA ⊆ R ◦ R and −1 that iB ⊆ R ◦ R . (5) Let R ⊆ A × B be a multivalued function (i) Let C ⊆ A. Prove that the restriction R|C×B ⊆ C × B is a multivalued function. (ii) Let now C * A, and let D be any set. Prove that RC×D ⊆ C × D is not a multivalued function. (iii) Prove that the restriction R|C×D ⊆ C × D is a multivalued function if, and only if, C ⊆ A ∧ (∀w)(w ∈ C → R[w] ∩ D 6= ∅). (6) Let R ⊆ A × B. Prove that −1 (i) R is a multi-valued function if, and only if, iA ⊆ R ◦ R; −1 −1 (ii) R is a correspondence if, and only if, iA ⊆ R ◦ R and iB ⊆ R ◦ R .

4.2.2 Partial Functions from a Set to a Set Deﬁnition 4.2.2 The relation R ⊆ A×B is a partial function from A to B if for every x, R[x] ⊆ B is an empty set, or is a singleton. Of course, R[x] = ∅ if x∈ / Dom(R) ⊆ A. As for any relation R ⊆ A × B we have Dom(R) ⊆ A, and Ran(R) ⊆ B, and the inclusions may be strong. For instance, we may have Dom(R) = ∅ (in which case Ran(R) = ∅ as well!).

When x ∈ Dom(R), we denote the only element of the set R[x] by R(x). We have therefore that R[x] = {R(x)}, and by definition (x, R(x)) is the only element of R having x as a first component. To distinguish partial functions as relations from a set to a set of a special type, we use specific notations to denote them. We often write

fR : A 9 B instead of (A, B, R), and set fR(x) = R(x) for x ∈ Dom(R). Often, when R is understood from the context, we write just f instead of fR.

Example 4.2.2 (1) The relation (A, A, iA) is a partial function. (2) The relations (∅,B, ∅) and (A, ∅, ∅) are partial functions. (3) Let A 6= ∅. The relation (A, B, A × B) is a partial function if, and only if, B is a empty or it is a singleton. (4) The relation (A, {x},R) is always a partial function. 2

48 Almost all operations on relations from a set to a set when applied to partial functions produce as a result partial functions!

Exercise 4.2.2 Let (A, B, R1) and (C,D,R2) be partial functions. Prove that composition, restriction, and intersection of these partial functions is a partial function. When it comes to inverse of a partial function, or to union of partial functions, the result may end up to be a relations which is not a partial function anymore. The conditions needed for these to be partial functions as well are described in the following exercise, but before that - a useful terminology:

Deﬁnition 4.2.3 The partial function f : A 9 B is called one-to-one if (∀x)(∀y)((x ∈ Dom(f) ∧ y ∈ Dom(f) ∧ x 6= y) → (f(x) 6= f(y))).

A partial function f : A 9 B is called onto if Ran(f) = B.

Exercise 4.2.3 (1) Prove that the inverse (A, B, R)−1 of a partial function is a partial function if, and only if, the partial function is one-to-one. (2) Let (A, B, R1) and (C,D,R2) be partial functions. Prove that the union (A ∪ C,B ∪ D,R1 ∪ R2) is a partial function if, and only if,

(R1)|(Dom(R1)∩Dom(R2))×(B∪D) and (R2)|(Dom(R1)∩Dom(R2))×(B∪D) are equal. The union is denoted in such a case also by

fR1 ∪ fR2 : A ∪ C 9 B ∪ D.

4.2.3 Functions from a Set to a Set Deﬁnition 4.2.4 Function from A to B is a relation R ⊆ A × B which is both a multivalued and a partial function. In other words, it is a relation such that (∀x)(x ∈ A → (R[x] is a singleton)).

In this case, for (x, y) ∈ R we write y = fR(x), or, when no danger of misunderstanding exists, just y = f(x). Such a relation is usually denoted by f : A → B. We deﬁne also the domain of f, Dom(f) := Dom(R) = A, and the range of f, Ran(f) := Ran(R). When y = f(x) we say that y is the value of f at x. In other words R ⊆ A × B is a function when it is a partial function which has Dom(R) = A. The range of f is described by {y | (∃x ∈ A)(y = f(x))}. Note that a function from A to B deﬁnes a rule assigning to every element x ∈ A a unique element f(x) ∈ B.

According to the deﬁnition of function, two functions are equal if, and only if, the relations they are deﬁned by are equal. That is, the two functions have to have the same domains, the same co-domains, and the same values at the elements of their domains: if f1 : A → B and f2 : C → D, then

(f1 = f2) ⇔ ((A = C =: X) ∧ (B = D =: Y ) ∧ (∀x)(x ∈ X → (f1(x) = f2(x)))).

Remark 4.2.1 In some books (such as the standard Calculus books), the deﬁnition of a function is ”a rule” assigning to an element of A a unique element of B without explaining what the rule is! Moreover, in such books the set

R = {(a, fR(a)) | a ∈ A} ⊆ A × B deﬁning the function is refered to as the graph of the function fR : A → B. In our approach, functions are rigorously deﬁned, and the ternminology properly used. We will not use the term graph of a function. 2

49 Example 4.2.3 (1) For every set A, the relation iA is a function (denoted by idA or by 1A). (2) Given a relation R ⊆ A × B, let the relation R˜ ⊆ A × P(B) be deﬁned by

R˜ := {(x, y) | (x ∈ A) ∧ (y = R[x]) }.

This relation is a function from A to P(B) which will be denoted by

R[·]: A → P(B). 2

Exercise 4.2.4 (1) Prove that, for every A, there is a unique function from ∅ to A. This function is denoted by ∅A : ∅ → A. Prove also that a function from A to ∅ exists if, and only if, A = ∅. (2) Prove that the relation R ⊆ A × B is a function if, and only if, (a) the family of sets

{R−1[y] | y ∈ B } ⊆ P(A) consists of pairwise disjoint elements, and (b)

∪|[R−1[y] | y ∈ B } = A.

Inversion, Restriction, Intersection, and Union of Functions from a Set to a Set We already know that the inverse relation of a class function need not be a class function. The same holds true in case of partial functions from a set to a set as well. In the latter case, the inverse (A, B, R)−1 of a partial function is a partial function if, and only if, the class relation R is one-to-one. The case of a function from a set to a set is even more restrictive when it comes to having the inverse relation a function as well:

Exercise 4.2.5 Let (A, B, F ) be a function. The inverse relation (B, A, F −1) is a function if, and only if, F is one-to-one, and for every y ∈ B the set F −1[y] is non-empty.

Deﬁnition 4.2.5 A function (A, B, F ) which is onto as a relation, that is which satisﬁes the property that (∀y)(y ∈ B → F −1[y] 6= ∅), is called a surjection. A function which is one-to-one as a partial function is called also injection. A function which is one-to-one (injection) and onto (surjection) is called a bijection. Obviously, bijections are one-to-one correspondences. We have that the inverse of a function from a set to a set is a function as well if, and only if, the function is a bijection.

Exercise 4.2.6 Prove that the function (A, B, F ) is a surjection if, and only if, Ran(F ) = B, if, and only if, ∪{F [x] | x ∈ A} = B. The other standard operations on relations we have been using such as restriction, intersection, and union have to also be cautiously applied to functions if we want to get a function as a result.

Deﬁnition 4.2.6 Let (A, B, F ) be a function, and let C and D be sets for which

C ⊆ A and {fF (x) | x ∈ C} ⊆ D.

The restriction (fF )|C×D : C → D is deﬁned by (C,D,F|C×D).

Exercise 4.2.7 Verify that the deﬁnition of restriction of a function above is correct: the result is a function indeed. Let f : A → B and g : C → D be two functions determined by (A, B, R) and (C,D,S) respectively. The intersection R ∩ S is a relation form (A ∩ C) to (B ∩ D) which, as we already know, is a partial function. We also know that Dom(R ∩ S) ⊆ A ∩ C, and that the inclusions may be strict.

50 Deﬁnition 4.2.7 In the notations above, the intersection f ∩ g is deﬁned when Dom(R ∩ S) = A ∩ C. In such a case, f ∩ g denotes the relation (A ∩ C,B ∩ D,R ∩ S).

The deﬁnition of union of two functions is very important, because it allows us to glue functions into new ones.

Deﬁnition 4.2.8 Let f : A → B and g : C → D be two functions determined by (A, B, R) and (C,D,S) respectively. The union f ∪ g is deﬁned only when the intersection f ∩ g exists. In such a case f ∪ g is determined by (A ∪ C,B ∪ D,R ∪ S).

Exercise 4.2.8 (1) Verify that the deﬁnition of union of two functions is correct: f ∪g is a function indeed having a domain A ∪ B. (2) Prove that if f : A → B and g : C → D are two functions such that A ∩ C = ∅, then f ∪ g does exist.

Remark 4.2.2 The constructions restriction and union of functions are often used in the Calculus course. Instances of restrictions of functions are the restriction of the domain of a trigonometric function in order to ”make it” invertible. It is common to restrict the domain and the co-domain of, say, the sine function from R to [−π/2, π/2] and to [−1, 1] in order to get an invertible function, and thus deﬁne sin−1. As for the union of functions, any function the students have seen being deﬁned piecewise (analytically on several intervals) is actually an example of union of (often more than two) functions. 2

Composition of Functions from a Set to a Set Functions, being (special types of) relations can be composed. The composition of two functions in general may only be a partial function, not a function. But when restricted to its domain, it is a function.

Exercise 4.2.9 Let R ⊆ A × B and S ⊆ C × D be functions. Let T = S ◦ R ⊆ A × D. Prove that T|Dom(T ) ⊆ Dom(T ) × D is a function.

There is an important particular case when the composition of two functions (as relations) is readily a function (without further restrictions of that composition). We say in such a case that either the functions are composable, or the corresponding relations are composable as functions.

Deﬁnition 4.2.9 The ordered pair of functions (f, g) are composable (as functions), if Ran(f) ⊆ Dom(g).

Exercise 4.2.10 (1) Prove that if (f, g) are composable, then their composition (as relations) g ◦ f is a function with domain Dom(f) and co-domain co − Dom(g) such that

(∀x ∈ Dom(f))((g ◦ f)(x) := g(f(x))).

(2) Prove that composition of functions is associative. That is, if (f, g) and (g, h) are composable pairs, then the pairs (g ◦ f, h) and (f, h ◦ g) are composable as well, and

h ◦ (g ◦ f) = (h ◦ g) ◦ f.

In particular, if f : A → B, g : B → C, and h : C → D are functions, then

h ◦ (g ◦ f) = (h ◦ g) ◦ f : A → D.

Note that (f, g) and (g, f) are composable if, and only if, Ran(f) ⊆ Dom(g) and Ran(g) ⊆ Dom(f).

51 Examples of Interesting Functions

Deﬁnition 4.2.10 Let f : A → B be a function, and let Y ⊆ B. The total pre-image of Y is the set {a ∈ A | f(a) ∈ Y }. This set is denoted by f −1(Y ).

Exercise 4.2.11 (1) Let the function f : A → B be given by the relation R ⊆ A×B. Prove that for −1 −1 every Y ⊆ B, f (Y ) = Ran((R )|Y ×A). This fact explains the notation for the total pre-image of Y . (2) Prove that, for every function f : A → B, the total inverse mapping Y 7→ f −1(Y ) deﬁnes a function P(B) → P(A).

Deﬁnition 4.2.11 Given a function f : A → B, we will denote by P∗(f): P(B) → P(A) sending Y ∈ P(B) to its total preimage under f, and will call it the total preimage function associated with f.

∗ A counterpart of P (f): P(B) → P(A) is the function P∗(f): P(A) → P(B) called the image function associated with f . Here is the deﬁnition1.

Deﬁnition 4.2.12 For a function f : A → B deﬁne P∗(f): P(A) → P(B) by

P∗(f)(S) = Ran(f|S) for any S ⊆ A.

∗ Some of the properties of the functions P∗(f): P(X) → P(Y ) and P (f): P(Y ) → P(X) associated with f : X → Y are collected in the following exercises.

∗ Exercise 4.2.12 (1) Prove that P (iA) = iP(A), P∗(iA) = iP(A). (2) Let (f, g) be composable functions, and let h = g ◦ f denote their composition. Prove that (i) the pair of functions (P∗(f), P∗(g)) is composable as well, and that P∗(h) = P∗(g) ◦ P∗(f). So, for a composable pair of functions (f, g) we have

P∗(g ◦ f) = P∗(g) ◦ P∗(f).

(ii) for the total inverse function P∗(h): P(D) → P(A) we have

(∀z ∈ P(D))(P∗(h)(z) = P(f)∗(co-Dom(f) ∩ P∗(g)(z)).

In particular, when co-Dom(f) = Dom(g), we have that (P(g)∗, P∗(f)) is a composable pair, and

P∗(h) = P∗(f) ◦ P∗(g).

∗ ∗ (iii) if f : A → B and g : B → C, then the pairs (f, g), (P∗(f), P∗(g)), and (P (g), P (f)) are composable, and ∗ ∗ ∗ P (g ◦ f) = P (f) ◦ P (g) P∗(g ◦ f) = P∗(g) ◦ P∗(f). (3) Let f : A → B be a function. Prove that ∗ (i) For every S ∈ P(A) we have S ⊆ (P (f) ◦ P∗(f))(S); ∗ (ii) For every T ∈ P(B) we have (P∗(f) ◦ P (f))(T ) ⊆ T .

1 ∗ −1 Neither P (f) nor P∗(f) is a traditional notation. The former is usually denoted by f while the latter is standardly denoted by f. The standard notations may easily lead to confusion: f −1 is also the notation for inverse function (to be introduced shortly), and f is the notation for . . . f. We suggest the new notations to avoid unwanted confusion caused by using the same mark for two diﬀerent objects ﬁrst, and, second, because these notation suggest ∗ that P∗ is a functor, and P is a co-functor from the category of sets Set to itself. (More about this - in the course Algebraic Structures.)

52 4.2.4 Most Important Functions from a Set to a Set Among all functions (from a set to a set), three types, which we have met already in this chapter, are most important: injection, surjection, and bijection. In this subsection, we are studying these functions via their compositions with other functions. We are showing also that any function can be expressed, in a natural way, as composition of an injection, bijection, and a surjection. This fact makes these functions so important.

Deﬁnition 4.2.13 Let f : A → B be a function. (1) f is called one-to-one or an injection if

(∀x, x0)((x, x0 ∈ A) ∧ (x 6= x0) → (f(x) 6= f(x0)));

(2) f is called onto or surjection if Ran(f) = B; (3) f is called a one-to-one correspondence or bijection if f is one-to-one and onto.

Example 4.2.4 The function 1A is a bijection. The empty map ∅B : ∅ → B is always an injection. ∅B is a surjection if, and only if, B = ∅. 2

Exercise 4.2.13 (1) (i) Let (f, g) be composable. Prove that if f and g are injections, then so is their composition g ◦ f. (ii) Let (f, g) be a composable pair. Show, by examples, that f and g surjections does not imply that g ◦ f is a surjection. (iii) Let co-Dom(f) = Dom(g). Prove that (f, g) is a composable pair such that if f and g are surjections (respectively, bijections), then so is g ◦ f. (2) Prove that f : A → B is a bijection if, and only if, there is a function g : B → A such that

g ◦ f = 1A f ◦ g = 1B.

[(⇐) Assume g exists. Need to show that f is ”one-to-one” and ”onto”. Suppose f(a1) = f(a2) for some a1, a2 ∈ A. Then g(f(a1)) = g(f(a2)), and therefore a1 = a2. This proves that f is one-to-one. Let now b ∈ B, and let a = g(b) ∈ A. We have f(a) = f(g(b)) = b, so B ⊆ Ran(f) ⊆ B, and therefore Ran(f) = B. This proves that f is onto. (⇒) Assume f is a bijection. Need to ﬁnd a g : B → A such that f ◦ g = 1B and g ◦ f = 1A. −1 To this end, let R ⊆ A × B be the relation deﬁning f: f = fR. Consider R ⊆ B × A. We have Dom(R−1 = Ran(R) = B, because f is onto. Also, (∀b ∈ B)(R−1[b] − singleton), because f is one-to-one. Therefore, R−1 is a function. Denote this function by g : B → A. Verify that this g does the job.] (3) Prove that, for f : A → B, if there is a function g : B → A as in (2), then this function is unique. This unique function is called the inverse function of f, and is denoted by f −1 : B → A. (4) Let f : A → B be a function. Prove that ∗ (i) P (f) ◦ P∗(f) = idP(A) if, and only if, f is an injection. ∗ [We know already that, for every S ⊆ A, S ⊆ P (f) ◦ P∗(f)(S). Prove that, for every S, the equality of the two sets happens if, and only if, f is one-to-one.] ∗ (ii) P∗(f) ◦ P (f) = idP(B) if, and only if, f is a surjection. ∗ [We know already that, for every T ⊆ B, P∗(f) ◦ P (f)(T ) ⊆ T . Prove that, for every T , the equality of the two sets happens if, and only if, f is onto.] ∗ (iii) the maps P∗(f): P(A) → P(B) and P (f): P(B) → P(A) are inverses of each other if, and only if, f : A → B is a bijection (so, f is invertible). Verify, in such a case, that

∗ −1 ∗ −1 P (f ) = P∗(f) and P (f) = P∗(f ).

(5) Given f : A → B, prove that (i) f is an injection if, and only if,

(∀g, h : C → A)((f ◦ g = f ◦ h) → (g = h));

53 (ii) f is a surjection if, and only if,

(∀g, h : B → C)((g ◦ f = h ◦ f) → (g = h)).

(6) Given A 6= ∅, and f : A → B, prove that f is an injection if, and only if, there is a g : B → A such that g ◦ f = 1A. [The function g is called a left inverse of f]. Show, by example, that such a function g may not be unique. [Let R ⊆ A × B be the relation defining f. Since f is injection, for every b ∈ B the set R−1[b] is either empty, or a singleton. Since A 6= ∅, fix a0 ∈ A. Define g : B → A as follows: g(b) = a −1 −1 if R [b] = {a}, and g(b) = a0 if R [b] = ∅. Verify that g ◦ f = 1A.]

Remark 4.2.3 After characterizing the bijections and the injections as in (2) and (6) above, one may be tempted to try to do the same for surjections as well. For this to be done, we need a new axiom! The next subsection explains the situation in detail.

Choice Functions and the Axiom of Choice

Te following concept turns out to play a central role in Set Theory.

Deﬁnition 4.2.14 Suppose X is a set such that ∅ ∈/ X. A function F : X → ∪X is called a choice function for X if (∀x)((x ∈ X) → (F (x) ∈ x)).

Note that for some sets, such as ∅ or any set containing ﬁnitely many elements (diﬀerent from ∅!), the existence of a choice function is not a problem! The only way to ensure the existence of a choice function in general is via an axiom.

Axiom 9 (Axiom of Choice) For every set X not containing ∅, there is a choice function

F : X → ∪X.

Exercise 4.2.14 (1) Can a set X with ∅ ∈ X have a choice function? (2) Given f : A → B, prove that f is surjection if, and only if, there is a g : B → A such that f ◦ g = 1B. [The function g is called a right inverse of f. Show, by example, that the right inverse of f may not be unique. [Let R ⊆ A × B be the relation deﬁning f, and consider R−1 ⊆ B × A. Assume B 6= ∅, and consider the set X = {w | (∃b ∈ B)(w = R−1[b]) }.

Since f is surjection, ∅ ∈/ X, and, according to the Axiom of Choice, there is a function

F : X → ∪X such that (∀x ∈ X)(F (x) ∈ x).

Deﬁne g : B → A as follows:

(∀b ∈ B)(g(b) = F (R−1[b])).

Verify that f ◦ g = 1B. What happens when B = ∅?] (3) Prove that ((2) ⇔ Axiom of Choice). (4) Let g be a left inverse, and h be a right inverse of the function f. Prove that g = h. (5) Prove that f is a bijection if, and only if, f has a unique right inverse, if, and only if, f has a unique left inverse.

54 4.2.5 Indexed Families of Sets Fulﬁlling a promise made earlier in the course, we are brieﬂy discussing the concept of indexed families of sets, and the related concept of multi-sets here.

We will make a serious use of indexed families later in the course, in the chapter of topology on the real numbers.

Deﬁnition 4.2.15 Indexed family of sets with index set Λ is a function F :Λ → X

The function F is the set of ordered pairs {(λ, F (λ)) | λ ∈ Λ} (considered as a subset of Λ × X of course!). We can denote by Xλ the set F (λ) for every λ ∈ Λ, and identify further every order pair (λ, F (λ)) with Xλ. With these new notations we have

F = {Xλ | λ ∈ Λ}.

The right-hand side of the equality above presents F as an indexed family of sets (each being an element of X), indexed by Λ. Observe also that if we consider the indexed family {Xλ | λ ∈ Λ} as a set, it is exactly Ran(F ). As a matter of fact, every set, as a collection of sets, can be represented as an indexed family of sets with index set - the set itself. Here is how this goes formally.

Example 4.2.5 Let Λ be a set. Its elements λ ∈ Λ, as we know, all are sets themselves. Let’s denote by Aλ the set λ. We have then that

Λ = {λ | λ ∈ Λ } = {Aλ | λ ∈ Λ },

and the set Λ is now considered as an indexed family of sets with index set Λ. 2

Exercise 4.2.15 Show that iΛ :Λ → Λ is the function which represents Λ as an indexed family with index set Λ.

Notice that when we consider a set as an indexed family of sets, such as Λ = {Aλ | λ ∈ Λ} above, then the family has the important property that its elements as members of the family are pairwise distinct: no Aλ = Aµ for distinct λ, µ ∈ Λ! In general though, for any family indexed by a set, repetition of elements are allowed! Remember that the notation Xλ, with Xλ = F (λ), is used as another name of the ordered pair (λ, F (λ)), and although Xλ = Xµ for some λ 6= µ may happen, as members of the indexed family the sets Xλ and Xµ are considered different! For instance, if f : A → 1 is a function, then for every a ∈ A, we should have f(a) = ∅, and accordingly, the indexed by A family of sets {Xa | a ∈ A} consists of identical elements: Xa = ∅. Therefore as a set the family is the singleton {∅}. So the family has many elements which are all identical as sets, but, due to the different indices on them, are distinct elements of the family. This flexibility the indexed families provide leads naturally to the concept of multi-sets and maps between them. We are not going any deeper in this here. The students will learn a bit more about the multi-sets in the course on Algebraic Structures.

A very important, for Calculus purposes, particular case of indexed families are the sequences of sets.

Definition 4.2.16 (1) An indexed family with indexing set N is called a sequence (of sets). (2) An indexed family with indexing set Λ is called infinite if there is an injection f : N → Λ, and is called finite otherwise.

Given an indexed family of sets F = {Xi | i ∈ I} we deﬁne the intersection of the family

∩F :=: ∩{Xi | i ∈ I} := {z | (∀i)(z ∈ Xi)}

55 (existing only when F 6= ∅), and the union of the family

∪F :=: ∪{Xi | i ∈ I} := {z | (∃i)(z ∈ Xi)}. Considering the indexed family F as a class function with domain I, that is, {(i, F(i)) | i ∈ I }, we have ∩ F = ∩ Ran(F) ∪ F = ∪ Ran(F).

In the notations Λ = {Aλ | λ ∈ Λ} introduced above, we can restate intersection and union of a set as intersection and union of an indexed family of sets as follows: for Λ 6= ∅,

∩Λ = ∩{Aλ | λ ∈ Λ} = {z | (∀λ ∈ Λ)(z ∈ Aλ)} as well as, for any Λ, ∪Λ = ∪{Aλ | λ ∈ Λ} = {z | (∃λ ∈ Λ)(z ∈ Aλ)}.

Exercise 4.2.16 (1) Prove that for any indexed family (non-empty if needed!) A := {Aλ | λ ∈ Λ} we have (∀λ ∈ Λ)(Aλ ⊆ ∪A ∧ ∩A ⊆ Aλ) (2) Prove that for every set B and for every indexed family A as in (1)

(∪A) ∩ B = ∪{Aλ ∩ B | λ ∈ Λ} (∪A) ∪ B = ∪{Aλ ∪ B | λ ∈ Λ};

(∩A) ∩ B = ∩{Aλ ∩ B | λ ∈ Λ} (∩A) ∪ B = ∩{Aλ ∪ B | λ ∈ Λ}. (3) Prove that for a set B and a family A as in (1)

B \ (∩A) = ∪{B \ Aλ | λ ∈ Λ} B \ (∪A) = ∩{B \ Aλ | λ ∈ Λ}.

4.3 Equivalence Relations 4.3.1 Basic Types of Relations on a Set Definition 4.3.1 Suppose R ⊆ A × A is a relation on A 6= ∅. (i) R is reflexive if (∀x)(x ∈ A → (x, x) ∈ R); (ii) R is irreflexive if (∀x)(x ∈ A → (x, x) ∈/ R); (iii) R is symmetric if (∀x, x0)((x, x0) ∈ R → (x0, x) ∈ R); (iv) R is asymmetric if (∀x, x0)((x, x0) ∈ R → (x0, x) ∈/ R); (v) R is anti-symmetric if (∀x, x0)(((x, x0) ∈ R ∧ (x0, x) ∈ R) → x = x0); (vi) R is transitive if (∀x, x0.x00)(((x, x0) ∈ R ∧ (x0, x00) ∈ R) → (x, x00) ∈ R); (vii) R is connected if (∀x, x0)((x, x0 ∈ A) → ((x, x0) ∈ R ∨ (x0, x) ∈ R ∨ x = x0)); (viii) R is trichotomous if it is connected, and for all x, x0 ∈ A exactly one of (x, x0) ∈ R, (x0, x) ∈ R, x = x0 holds true. Notice that the negation of reflexive is not irreflexive. Neither is any of asymmetric and antisym- metric the negation of symmetric!

Exercise 4.3.1 Can in this deﬁnition A be the empty set? If yes, give proof, and if no, give arguments why not.

Example 4.3.1 (1) The (class-) relation ∈ is irreflexive (by an axiom!). (2) The relation ≤ on N is reflexive, anti-symmetric, transitive, and connected. It is not trichotomous. (3) The relation < on N is irreflexive, asymmetric, transitive, connected, and trichotomous. 2

56 4.3.2 Equivalence Relations v/s Partitions on a Set, Quotient Maps In this subsection, we are introducing a very important type of relation on a set: the equivalence relation. This relation forms a convenient tool to study and construct sets. We will have several occasions to use this type of relations in this course.

Definition 4.3.2 The relation R ⊆ A × A is an equivalence relation on A if R is reflexive, symmetric, and transitive. A useful way to characterize the equivalence relations is by using the function R[·]. Recall that, for any relation R ⊆ A × B, we defined a function R[·]: A → P(B) given by x ∈ A 7→ R[x] ∈ P(B). In general, R[x] may be the empty set, or for some x, x0 ∈ A with R[x] 6= R[x0] we may have R[x] ∩ R[x0] 6= ∅. Indeed, let R be the relation ≤ on N. In this case, R[1] 6= R[2], but R[1] ∩ R[2] = {n | 2 ≤ n}= 6 ∅. Take as another example R to be the inverse relation of < on N. In this case R[0] = ∅. Well, if R is an equivalence relation on A, and A 6= ∅, the function R[·]: A → P(A) behaves much more gently:

Theorem 4.3.1 The relation R on A 6= ∅ is an equivalence relation if, and only if, (i)(∀x)((x ∈ A) → (x ∈ R[x])) and (ii) (∀x, x0)((R[x] 6= R[x0]) → (R[x] ∩ R[x0] = ∅)). Proof. (⇒) Assume that R is an equivalence relation on A. Since R is reflexive, for every x ∈ A we have that (x, x) ∈ R, so that x ∈ R[x]. This proves item (i). We will prove item (ii) now. Assume x00 ∈ R[x] ∩ R[x0]. We want to show that R[x] = R[x0]. Note first that, since R is symmetric, and transitive, x ∈ R[x] and x0 ∈ R[x]. Indeed, we have (x, y) ∈ R ∧ (x0, y) ∈ R, and so (x, y) ∈ R ∧ (y, x0) ∈ R which gives that (x, x0) ∈ R, and therefore x0 ∈ R[x]. Analogously, x ∈ R[x0]. But x0 ∈ R[x] implies that R[x0] ⊆ R[x]. That is, because (x0, y) ∈ R ∧ (x, x0) ∈ R implies that (x, y) ∈ R, and therefore y ∈ R[x]. In a similar fashion R[x] ⊆ R[x0]. The two inclusions of equivalence classes prove that R[x] = R[x0]. Item (ii) is proved. (⇐) We have to prove now that if R is a relation on a non-empty set A and if it has the properties of items (i) and (ii), then R is reflexive, symmetric, and transitive. Item (i) says exactly that for every x ∈ A,(x, x) ∈ R. So R is reflexive. Now, proving the symmetry of R, let (x, x0) ∈ R. This means that x0 ∈ R[x0] ∩ R[x], and by (ii) it follows that R[x] = R[x0], but then x ∈ R[x0] which means that (x0, x) ∈ R. In a similar fashion, for the transitivity, we have (x, x0) ∈ R and (x0, x00) ∈ R imply that R[x] = R[x0] and R[x0] = R[x00]. So, the three sets are equal R[x] = R[x0] = R[x00]. In particular, R[x] = R[x00], and (x, x00) ∈ R. The relation R is an equivalence relation. 2

Exercise 4.3.2 Is this theorem true for A = ∅? Give reasons for your answer.

Corollary 4.3.2 If R is an equivalence relation on a non-empty set A, then the set {R[x] | x ∈ A } consists of non-empty pair-wise disjoint sets such that ∪{R[x] | x ∈ A } = A. Proof. Follows directly from the theorem above. 2

Exercise 4.3.3 Prove this Corollary. This corollary motivates the following deﬁnition.

Deﬁnition 4.3.3 Let A be a non-empty set, and let B ⊆ P(A). Then, B is a partition of A if (1) (∀x)((x ∈ B) → (x 6= ∅)); (2) (∀x)(∀x0)(({x, x0} ⊆ B ∧ x 6= x0) → (x ∩ x0 = ∅)); (3) ∪B = A.

57 By Corollary 4.3.2, we know that every equivalence relation R of a non-empty set A deﬁnes a partition of A. This partition is denoted by BR. The fact of the matter is that, conversely, every partition of a non-empty set A deﬁnes a relation on A, which turns out to be an equivalence relation. Namely,

Deﬁnition 4.3.4 Suppose B is a partition of the non-empty set A. Deﬁne RB ⊆ A×A by declaring that for x, x0 ∈ A 0 0 (x, x ) ∈ RB if (∃w)((w ∈ B) ∧ ({x, x } ⊆ w).

Proposition 4.3.3 For any partition B of a non-empty set A, the relation RB on A is an equivalence relation.

Proof. An easy check that RB is reﬂexive, symmetric and transitive. 2

Exercise 4.3.4 Prove this Proposition. The next theorem reveals the fact that the concepts of equivalence relations on a non-empty set and of partitions of that set are identical. Recall that the relations on a set A comprise a set - the power set P(A × A) of A × A. Therefore, the equivalence relations on A form a set as well (verify that! That is write down a formula deﬁning the equivalence relations among all relations on A). Denote this set by EquivRel(A). On the other hand, the partitions of A are a sub-collection in P(A), so they form a set as well (write the formula here, too!). Denote that set by P artit(A). The theorem and the proposition above provide us with two maps between EquivRel(A) and P artit(A)

ϕA : EquivRel(A) → P artit(A) R 7→ ϕA(R) := BR and ψA : P artit(A) → EquivRel(A) B 7→ ψA(B) := RB.

Exercise 4.3.5 Describe P artit(∅) and EquivRel(∅).

Theorem 4.3.4 Let A be a non-empty set. The maps ϕA and ψA are inverses of each-other. In

other words, for any equivalence relation R on A, we have RBR = R, and, for any partition B of A,

we have BRB = B.

Proof. Let R be an equivalence relation on A. To prove that RBR = R ⊆ A × A we have to show that 0 0 0 (∀x, x ∈ A)((x, x ) ∈ R ↔ (x, x ) ∈ RBR ). 0 0 0 But, (x, x ) ∈ R is equivalent to having R[x] = R[x ] ∈ BR. This latter implies that (x, x ) ∈ RBR . 0 00 0 00 Conversely, if (x, x ) ∈ RBR , then there is an equivalence class R[x ] ∈ BR such that x, x ∈ R[x ]. That is: (x00, x) ∈ R and (x00, x0) ∈ R. But since R is an equivalence relation, (x, x0) ∈ R. This

completes the proof of RBR = R.

The proof of the coincidence B = BRB for any partition B of A is left as an exercise. 2

Exercise 4.3.6 Prove, in the notations of the previous theorem, that B = BRB for any partition B on A. The partition determined by an equivalence relation of a non-empty set A has a special name, and numerous uses in math.

Deﬁnition 4.3.5 Let R be an equivalence relation on A 6= ∅. The partition BR is called the quotient set of A with respect to R, and is denoted by A/R. The function R[·]: A → P(A) deﬁnes a map πR : A → A/R called the quotient map. Note that πR is a surjection.

0 0 By the above, we know that (x, x ) ∈ R if, and only if, πR(x) = πR(x ).

58 Application of Equivalence Relations on a Set Equivalence relations play a very important role in math. We will use equivalence relations to construct the basic number systems (integers, rational, and real numbers) in the following chapters. In this sub-section we address some promises made as well as some constructions already discussed earlier in the course, and which can be treated by using equivalence relations.

As we mentioned earlier, the three types of set maps, injection, surjection, and bijection, are the most important ones. This is due to the fact that any set map can be expressed, in a natural way, as a composition of surjection, followed by a bijection, and an injection. The following exercise discusses some properties of equivalence relations on a set. Item (4) of the exercise, in particular, suggests how to prove this fact above.

Exercise 4.3.7 (1) This exercise proves in effect that every function is a composition of an injection and a surjection both defined in a natural way. Let f : A → B be a function with A 6= ∅. Define a relation Rf ⊆ A × A by

0 0 0 (∀x, x )((x, x ) ∈ Rf if (f(x) = f(x ))).

Verify that Rf is an equivalence relation on A. Prove further that there is a unique function ϕ :

A/Rf → B such that f = ϕ ◦ πRf . Prove that the function ϕ is an injection.

f A / B <

πR ∃!ϕ f " < A/Rf

(2) The fact proved in this exercise describes a universal property satisﬁed by the quotient map πR : A → A/R. The importance of this, and other universal properties satisﬁed by set maps, will be revealed in the course of Algebraic Structures where they play important roles. Let R be an equivalence relation on A 6= ∅, and let πR : A → A/R be the corresponding quotient map. Prove that, for every function f : A → B such that

(∀x, x0)((x, x0) ∈ R → f(x) = f(x0)),

there exists a unique ϕ : A/R → B such that f = ϕ ◦ πR.

f A / B =

πR ∃!ϕ ! A/R

(3) This exercise is a ﬁner version of the previous two. The set up is as follows. Suppose R is an equivalence relation on A 6= ∅, and S is an equivalence relation on B 6= ∅. Suppose further that f : A → B is (R − S)-equivariant, that is

(∀x, x0 ∈ A)((x, x0) ∈ R → (f(x), f(x0)) ∈ S). ˜ ˜ Prove that there is a unique map f : A/R → B/S such that f ◦ πR = πS ◦ f

f A / B

πR πS ∃!f˜ A/R / B/S

59 (4) In this exercise, the proper claim about representing a set map as a composition of surjection, bijection, and injection in a natural way is stated. Prove it. Let f : A → B be any set map. Let further π : A → A/R and i : Ran(f) → B be the quotient Rf f f map, and the inclusion map respectively. Prove that there is a unique set map ϕ : A/Rf → Ran(f) such that f = i ◦ ϕ ◦ π . Moreover, ϕ is a bijection. f Rf

f A / B O π Rf if ∃!ϕ A/Rf / Ran(f) ∼= The quotient map π and the inclusion map i are defined ”in a natural way” by the map f. The Rf f map ϕ is unique. It is in this sense that the presentation of f as a composition of the named maps is natural. (5) Let R be (any) relation on A 6= ∅. Consider the function R[·]: A → P(A). By (1) above, (the function) R[·] defines an equivalence relation RR[·] on A. Explain this equivalence relation in terms of R. Who is the quotient set A/RR[·]? The same question when R is an equivalence relation on A. We discussed earlier in the Notes a description of the concept of multi-set. We used for this description the notion of an indexed family of sets. The next exercise offers another description of multi-sets based on the notion of equivalence relation. It suggests to prove also that the two approaches, by indexed families of sets and by equivalence relation, are equivalent.

Exercise 4.3.8 Let R be an equivalence relation on the set A. Let’s agree to call two elements of A identical if they are R-related. This will be our new model of a multi-set. Consider the quotient map πR : A → A/R associated with R as a function, and consider the indexed family of sets it defines. This indexed family describes a multi-set as discussed earlier in this chapter. Prove that the two descriptions are equivalent in the following sense. (i) The description of multi-set using equivalence relation R on a set A determines in a natural way an indexed family of sets A, indexed by A, i.e., πR : A → A/R, and (ii) the description of a multi-set using indexed family of sets A, indexed by a set A, i.e., f : A → X, determines an equivalence relation on A in such a way that (*) if we begin with an equivalence relation R on A, and define the indexed family of sets, the equivalence relation on A the latter defines is R, and (**) if we begin with an indexed family of sets f : A → X, and define the equivalence relation Rf on A as in (ii), the indexed family of sets associated with the function π : A → A/R is naturally Rf f isomorphic to the original one. This means that the unique ϕ : A/Rf → X which exists by item (1) of the previous exercise is a one-to-one with Ran(ϕ) = Ran(f)

f A / X < 1−1 π ∃!ϕ Rf " A/Rf

4.4 Orders

Orders on a set reﬂect preferences on some elements over others.

4.4.1 Different Types of Orders on a Set Definition 4.4.1 Let R be a relation on the set A. This relation is called (1) a pre-order on A if R is reflexive and transitive;

60 (2) a partial order on A if R is reﬂexive, anti-symmetric, and transitive; (3) a strict partial order on A if R is irreﬂexive and transitive; (4) a total order if R is a connected partial order; (5) a strict total order on A if R is a connected strict partial order; (6) dense order on A if R is a total order such that

(∀x 6= y ∈ A)(∃z ∈ A \{x, y})(((x, z) ∈ R ∧ (z, y) ∈ R)));

(7) to have the least upper bound property if

(∀B ⊆ A)((B 6= ∅ ∧ bounded above in A) → (B has a least upper bound in A));

(8) to have the greatest lower bound property if

(∀B ⊆ A)((B 6= ∅ ∧ bounded below in A) → (B has a greatest lower bound in A));

(9) a continuous order on A if R is dense, and has the least upper bound property. (10) a well order on A if R is a partial order and

(∀B ⊆ A)((B 6= ∅) → (B has a least element w.r.t. R)) where bounded below, bounded above, least element, and least upper bound are deﬁned below.

Example 4.4.1 (1) The relation ≤ on N is a well order. This order is not dense, and has the least upper, and the greatest lower bound properties. Indeed, the relation is a well order due to Chapter 3 of these Lecture Notes. It is not dense, because there is no natural number strictly between n and n + 1 for every n ∈ N. The least upper bound property: Let ∅= 6 A ⊆ N be bounded above; so there is m ∈ N such that for every n ∈ A, we have n ≤ m; but then m + 1 ∈/ A, and

B = {m | m ∈ N ∧ m is a l.u.b. of A}= 6 ∅; so, B has a least element: s; then s is an u.b. of A; it is the a l.u.b. as well, because if there were s0 < s such that s0 is an upper bound of A, then s0 would necessarily be in B, and, for the inequality above, s wouldn’t be a least element of B. By the way, s is necessarily in A! Do you see this? In a similar way, one proves that the relation ≤ has the greatest lower bound. Do that as an exercise. (2) The relation < on N is a strict total order. It is not dense, and has the least upper and the greatest lower bound property. The arguments here are similar to the ones of item (1) of this exercise. (3) The relation ⊆ on P(A), where A is a set, is a partial order. It has the least upper and the greatest lower bound properties. Checking out the that the deﬁnition of partial order is satisﬁed is straightforward. Let A ⊆ P(A) be a non-empty family of subsets of A. The least upper bound is, obviously, ∪ A. The greatest lower bound is, equally obviously, ∩ A. Check this out! (4) The relation ⊂ on P(A) is a strict partial order. It has the least upper and the greatest lower bound properties. The argument here is similar to the one in item (3) of this exercise. 2

4.4.2 Minimal Elements, Maximal Elements, and All That Stuﬀ Deﬁnition 4.4.2 Let R be a partial order on A. (1) m ∈ A is a minimal element w.r.t. R if (∀x ∈ A)((x, m) ∈ R → x = m); (2) M ∈ A is a maximal element w.r.t. R if (∀x ∈ A)((M, x) ∈ R → x = M); (3) m ∈ A is a least/minimum element w.r.t. R if (∀x ∈ A)((m, x) ∈ R); (4) M ∈ A is a greatest/maximum element w.r.t. R if (∀x ∈ A)((x, M) ∈ R). In what follows B ⊆ A. (5) u ∈ A is an upper bound (u.b.) of B w.r.t. R if (∀z ∈ B)((z, u) ∈ R); B is called bounded

61 above in A if it has an u.b. w.r.t. R; (6) l ∈ A is a lower bound (l.b.) of B w.r.t. R if (∀z ∈ B)(((l, z) ∈ R)); B is called bounded below in A if it has a l.b. w.r.t. R; (7) u ∈ A is a least upper bound (l.u.b.) of B w.r.t. R if u is the least element, w.r.t. R, of the set of upper bounds of B in A. (8) l ∈ A is a greatest lower bound (g.l.b.) of B w.r.t. R if l is the greatest element, w.r.t. R, of the set of lower bounds of B in A. (9) B ⊆ A is an initial segment in A w.r.t. R if (∀y)((y ∈ B) → ((∀x ∈ A)((x, y) ∈ R → x ∈ B))). 10) B ⊆ A is a ﬁnal segment in A w.r.t. R if (∀y)((y ∈ B) → ((∀x ∈ A)((y, x) ∈ R → x ∈ B))).

Example 4.4.2 (1) It is important in the deﬁnitions of least upper and of greatest lower bound property to have B 6= ∅. Indeed, if A = N, and if the relation is ≤ or <, then every natural number is an upper and a lower bound of B = ∅ ⊆ N Therefore, the least upper bound of B in this case is 0 ∈ N while the greatest lower bound does not exist. (2) Every natural number is an initial segment of (N, <). Indeed, let n ∈ N, and m ∈ n. For and s ∈ N such that s < m, we have, by the transitivity if <, that s < n as well. By the very deﬁnition of <, we have that s ∈ n. Similar statement holds true for the relation ≤ as well. (3) Consider A = P(X) with the relation ⊆ or (. Let Y ⊆ X. Then P(Y ) is an initial segment of A. Indeed, if Z ∈ P(Y ), then Z ⊆ Y , and if W ⊆ Z, respectively W ( Z, then W ⊆ Y , respectively W ( Y , and therefore W ∈ P(Y ). The initial segment P(Y ) has Y as its maximum element, i.e. the element. The set P(Y ) \{Y } is also an initial segment in P(X) with the aforementioned relations (check this out), it doesn’t have a maximum element, but still has a least upper bound, the set Y itself. (4) Any set is a partially ordered set w.r.t. inclusion among its elements (these are all sets!). With this order, every A 6= ∅ has greatest lower bound, given by ∩ A. Any set A, empty or not, has a least upper bound, given by ∪ A. 2

Exercise 4.4.1 (1) Prove that if R is a partial order on A, then the relation R0 on A deﬁne by (∀x, x0 ∈ A)((x, x0) ∈ R0 if ((x, x0) ∈ R) ∧ (x 6= x0)) is a strict partial order on A. Conversely, prove that if R is a strict partial order on A, then the relation R0 on A deﬁned by (∀x, x0 ∈ A)((x, x0) ∈ R0 if ((x, x0) ∈ R) ∨ (x = x0)) is a partial order on A. (2) Prove that if R is a pre-order on A, then R ∩ R−1 is an equivalence relation on A. (3) Prove that the least upper bounds and the greatest lower bounds are unique whenever they exist. (4) Prove that if R is a well order on A, then R is a total order which has the least upper bound, and the greatest lower bound properties. (5) Prove that the relation ≤ on N is not continuous. (6) Prove that the initial segments of N w.r.t. either ≤ or < are the elements of N and the set N itself. (7) Let B be an initial segments of P(X) w.r.t. either ⊆ or (. Prove that if B has maximum element Y in P(X), then B = P(Y ). (7’) There are many other types of initial segments in P(X). Let {Xλ | λ ∈ Λ } be a family of subsets of X. Prove that B = ∪ {P(Xλ) | λ ∈ Λ } is an initial segment in P(X). As matter of fact, any initial segment in P(X) is a set of this form: a union of power sets of a family of sub-sets of X. (8) Let R be a relation on A 6= ∅ which is a pre-order, a partial order, a total order, or a strict order (partial or total). Consider the function R[·]: A → P(A). Explain, in terms of R, the equivalence relation, RR[·], on A determined by that function. Who is the quotient set A/RR[·] (in all cases of R above)?

62 4.4.3 Axiom of Choice, Well Orders, Zermelo’s Theorem, Zorn’s Lemma In this subsection, we are discussing two theorems which are equivalent to the Axiom of Choice: Zermelo’s Theorem and Zorn’s Lemma. The importance of these theorems stems from their wide applicability in diﬀerent situations in mathematics.

The formulation of Zorn’s Lemma needs some preliminaries.

Deﬁnition 4.4.3 Suppose R is a partial order on A, x, x0 ∈ A, and B ⊆ A. 0 0 (1) The elements x, x are (R-)comparable if R|{x,x0} is a total order on {x, x }; (2) The subset B is a chain/an R-chain if R|B is a total order on B. Consider the following propositions

(Zermelo) Every non-empty set is well orderable.

(Zorn) If R is a partial order on A such that every R-chain B has an upper bound in A, then A has a maximal element.

The German set theorist Ernst Friedrich Ferdinand Zermelo proved, in 1904, the Zermelo’s Theorem to the eﬀect that the Axiom of Choice is equivalent to (Zermelo). The German algebraist Max Zorn proved, in 1935, the Zorn’s Lemma that the Axiom of Choice is equivalent to (Zorn). Thir- teen years earlier, in 1922, this theorem was proved by the Polish set theorist Kazimierz Kuratowski.

Problem. Using the theory of Axioms 1-8, prove that the Axiom of Choice, (Zermelo), and (Zorn) are equivalent propositions. [Sketch of a Proof. (Zorn)⇒(Zermelo). Let X 6= ∅ be given. Assuming (Zorn), we want to show that there is an order R on X which is a well order. The idea of the proof is to consider all subsets of X which have well orders on them, and choose a largest one among them. Turns out, any such largest one has necessarily to be X. So, consider the set

Σ = {(Y,S) | (Y ⊆ X) ∧ (S ⊆ Y × Y − well order)}.

For every (Y,S) ∈ Σ and for y ∈ Y , call initial segment of y in Y the set

P redS(y) = {w ∈ Y | (w, y) ∈ S ∧ w 6= y }.

Deﬁne a relation on Σ as follows

0 0 0 0 (Y,S) (Y ,S ) if (Y ⊆ Y ) ∧ (S = S |Y ) ∧ (∀y ∈ Y )(P redS(y) = P redS0 (y)).

This relation is a partial order on Σ (Check this out!). We are verifying next that the premise of (Zorn) holds true for (Σ, ). To this end, let Λ = {(Yλ,Sλ) |λ ∈ Λ } be a chain in (Σ, ). We want to show that Λ has an upper bound in Σ. Let Y = ∪{Yλ|λ ∈ Λ}. Observe that

(∀λ ∈ Λ)( Sλ ⊆ Yλ × Yλ ⊆ Y × Y ),

so that the set S = ∪{Sλ | λ ∈ Λ } ⊆ Y × Y deﬁnes a relation on Y .

Exercise 4.4.2 Prove that the relation S can be described also by (y1, y2) ∈ S if, and only if, there is a λ ∈ Λ such that (y1, y2) ∈ Sλ. It is straightforward now that S is a total order on Y . This order is actually a well order on Y .

Indeed, let ∅= 6 Z ⊆ Y , and let z0 ∈ Z. By the deﬁnition of Y , there is a λ0 ∈ Λ such that z0 ∈ Yλ0 . We claim that P red (z ) = P red (z ). S 0 Sλ0 0

63 The inclusion P red (z ) ⊆ P red (z ) follows from S ⊆ S. For the opposite inclusion, let Sλ0 0 S 0 λ0 w ∈ P redS(z0), that is, (w, z0) ∈ S. By the deﬁnition of S, there is a µ ∈ Λ such that (w, z0) ∈ Sµ.

Since Λ is a chain in (Σ, ), either (Yλ0 ,Sλ0 ) (Yµ,Sµ) or (Yµ,Sµ) (Yλ0 ,Sλ0 ). In either case, by the deﬁnition of , we have that w ∈ P red (z ). This ﬁnishes the proof of the claim. Sλ0 0 Returning to proving that S is a well order, we are showing that Z has a least element. If z0 is a least element, we are done. If not, P redS(z0) 6= ∅, and therefore

∅= 6 P red (z ) ∩ Z = P red (z ) ∩ Z ⊆ Y , S 0 Sλ0 0 λ0 and therefore P redS(z0) ∩ Z has a least element y0. Since S is a total order on Y , the element y0 is a least element of Z as well (Prove that!). This ﬁnishes the proof that (Y,S) is a well order. This implies that (Y,S) ∈ Σ, and that it is an upper bound for Λ. Having veriﬁed that the premise of (Zorn) holds true for (Σ, ), we conclude that there is a maximal element (Y0,S0) ∈ Σ. Our last claim is that Y0 = X. This follows from the next Exercise.

0 0 0 0 Exercise 4.4.3 Let Y0 $ X, and let z ∈ X \ Y0. Deﬁne a relation S on Y = Y0 ∪ {z } as follows

0 0 0 0 S = S0 ∪ {(y, z ) | y ∈ Y0} ∪ {(z , z )}.

0 0 0 0 Prove that (Y ,S ) ∈ Σ, and that (Y0,S0) (Y ,S ).

0 0 Conclude from this Exercise that Y0 = Y which is an absurd by the construction of Y . This ﬁnishes the proof of (Zorn)⇒Zermelo).

(Zermelo)⇒(Axiom of Choice). Given X such that ∅ ∈/ X we want, assuming (Zermelo), to prove that there is a choice function F : X → ∪X. According to (Zermelo), every element of X is well orderable. Consider, for every x ∈ X, a well order Sx on it, and denote by x0 ∈ x the corresponding least element. The function F : X → ∪X is readily deﬁned: (∀x ∈ X)(F (x) = x0). This completes the proof of (Zermelo)⇒(Axiom of Choice).]

Ideally, we have to prove also that (Axiom of Choice)⇒(Zorn) in order to complete the proof of the claim in the Problem. But this implication is really tricky to prove, if no deeper theory of ordinal numbers is used. We leave it to the really diligent students to supply a proof for themselves.

We end this subsection proving that (Zermelo)⇒(Zorn). Our argument (borrowed from Aluﬃ’s book ”Algebra, Chapter 0”) uses recursion in a signiﬁcant way.

Proof of (Zermelo)⇒(Zorn). Given a partially ordered set (X,R) which satisfies the premise of (Zorn), we have to prove, using (Zermelo), that (X,R) has a maximal element. By (Zermelo), there is a well order S on X. We are constructing next a chain in (X,R). We are doing that using a function defined recursively, with respect to the order S, on X. To this end, let x0 ∈ X be the S-least element of X, and define F (x0) = {x0}. We define F (x) for x ∈ X, x 6= x0 the following (recursive) way. Suppose F has been defined on P redS(x), and consider the set

Σx = {x} ∪ {F (y) | y ∈ P redS(x) }.

If Σx is an R-chain, then define F (x) = {x}. If Σx is not an R-chain, define F (x) = ∅. We claim first that Dom(F ) = X. Indeed, if not, then

∅= 6 Y = X \ Dom(F ) ⊆ X,

and since S is a well order, there is an S-least element y0. Obviously, P redS(y0) ∩ Y = ∅, and therefore F is defined on P redS(y0). But then it is defined for y0 as well, which means that y0 ∈/ Y which contradicts the definition of y0 as the least element of Y . This proves that Dom(F ) = X. The chain in (X,R) is defined by Λ = ∪Ran(X) (verify as an exercise that Λ is indeed an R-chain in X). Since (X,R) satisfies the premise of (Zorn), Λ has an R-upper bound x0 ∈ X. We finish our

64 proof showing that x0 is an R-maximal element of X. Observe ﬁrst that any upper bound y of Λ belongs to Λ. That’s because Σy ⊆ Λ ∪ {y}, which implies that Σy is an R-chain, and consequently, F (y) = {y}. Let now (x0, x00) ∈ R. Then x00 is an upper bound of Λ as well, and by the just proven property of the upper bounds of Λ, x00 ∈ Λ. But then (x00, x0) ∈ R as well, and since R is anti-symmetric, x0 = x00. This completes the proof of (Zermelo)⇒(Zorn). 2

4.4.4 Concluding Remarks On the Axioms we Introduced In the previous three chapters, we presented a development of Set Theory based on axioms. In our presentation, we haven’t introduced one axiom (the Axiom of Replacement, and have used a weaker version of another one: the Axiom of Foundation. Both axioms are not needed for our course. The former is useful when working with Class relations and with Class functions. These are covered in the more complete version of these lecture notes. The latter we discuss brieﬂy here. The Axiom of Foundation states that every non-empty set is well founded, that is

(∀X)(X 6= ∅ → (∃x)(x ∈ X ∧ x ∩ X = ∅)).

Our Axiom 7 follows from the Axiom of Foundation applying it to the singleton {x}. Indeed, according to the Axiom of Foundation, {x} has an element (there is only one element in it: x) which intersects with the set emptily, that is, x ∩ {x} = ∅. The last relation means that x∈ / x. The theory of Axioms 1 through 8, with the Axiom of Foundation instead Axiom 7,and with the Axiom of Replacement is the Zermelo-Fraenkel axiomatic Set Theory (ZF). The theory of Axioms 1 through 10 is the aximatic Set Theory of Zermelo-Fraenkel with Axiom of Choice (ZFC). The axioms of ZFC are generally accepted to deal with sets.

As promised in Remark 4.1.2, using the Foundation Axiom we can answer the question if

((x, y), z) = (x, (y, z))

for some sets x, y and z. Indeed, the equality of the two sets is equivalent to

x = (x, y) and z = (y, z)

or, equivalently to x = {{x}, {x, y}} and z = {{y}, {y, z}}. Since x ∩ {x} = ∅, the Foundation Axiom is satisﬁed by the set x. The same Axiom applied to the set z tells us that y∈ / z. Contrapositively, if y ∈ z, the equality z = {{y}, {y, z}} is impossible. Therefore, ((x, y), z) = (x, (y, z)) is also impossible in such a case. As a matter of fact, the latter equality is never possible, even when y∈ / z, but to prove this we need theory going far beyond the scope of our these Lecture Notes, and which is presented in any decent course on Set Theory.

What Is Coming Next in These Lecture Notes

We already know that the relation ≤ on N has least upper and greatest lower bound properties, but is not dense (verify that!), so - not continuous. In the next part of the course, we will construct, using set theory, the smallest ﬁeld extension of N which will inherit the relation ≤ from N as a continuous relation. This extension is needed for the problems in Calculus related to limits of sequences and functions.

We will ﬁrst construct the set of integers Z for the purposes of arithmetic. It will inherit the relation ≤ from N. On Z the relation ≤ will still not be a dense relation, having both types of bounds. We will construct next the set of rational numbers, Q, again for arithmetic reasons. This set will be a ﬁeld, the relation ≤ will there be dense, but with no least upper and greatest lower bound properties, and therefore - not continuous. Finally, we will construct the set of real numbers

65 R. The relation ≤ will have the needed property of being continuous there.

The sets N, Z, Q and R are completely diﬀerent sets, not sharing any elements. But we will also construct natural injections N → Z, Z → Q, and Q → R which will allow us to enjoy the usual picture we are familiar with since preschool:

N ⊂ Z ⊂ Q ⊂ R.

66 Chapter 5

Construction of the Standard Number Systems

In this chapter, we are giving the construction of the sets of integer numbers, Z, rational numbers, Q, and real numbers, R. For this we use set theoretical tools. The idea is to construct the number systems in such a way that they have two operations - addition and multiplication, and order - all inherited from the set of natural numbers N. In technical terms, the integers form a ring, while the rationals and the reals form fields (that is, in them one can divide by non-zero elements). That’s why they are better, from point of view Algebra, than the naturals. On the other hand, the field of reals is better, from point of view of Calculus, than the field of rationals, because in it,contrary to the rationals, limits of sequences exist. All these sets are ”essentially unique”, in a sense described below, with their properties. For instance, the set of integers is the smallest ring containing the naturals, the set of rationals is the smallest field containing the integers as a sub-ring, and the set of real numbers is the smallest ordered and complete field which contains the rational numbers as a dense sub-field.

5.1 The Integers 5.1.1 Motivations; Deﬁnition of the Set of Integers

In doing arithmetic in N, one is led to consider equations of the type mX + n = s

where m, n, s ∈ N. Assuming, for simplicity, that m = 1, one needs to be able to solve the equation X + n = s.

Within N this can be solved only if n ≤ s. To get a solution without this restriction, we need to expand the set of natural numbers so that all the solutions to that equation for s < n be in it. Call the expansion X. So, we have N ⊆ X. The set X has to be constructed in such a way that it retains most of the properties of N. In particular, two operations, addition and multiplication, and a relation, ≤, have to be deﬁned on X so that their restriction to N coincides with the usual operations and the relation on N. The operations on X have to be commutative and associative, they need to have 0, 1 ∈ N ⊆ X as respective neutral elements, and the multiplication should distribute over the addition. Also, notice that in order to be able to solve the (simpliﬁed) equation above, it is enough to have that for every n ∈ N, there is a y ∈ X such that n + x = 0, and therefore this latter property has to be a feature of X as well. Furthermore, the relation ≤ has to be a total order on X with the properties of additive and multiplicative cancellation. Obviously, any such expansion X of N will allow us to solve that equation. The set of integers is

67 determined as being a minimal such expansion. The minimality can be described by the property that for every y ∈ X either y ∈ N, or there is an n ∈ N such that y + n = 0. After these preliminary considerations we are ready to give a deﬁnition of the set of integers.

Deﬁnition 5.1.1 A set X with two operations +X : X × X → X and ·X : X × X → X is called a set if integers if (1) (X, +X ) is a commutative group. That is, +X is commutative, associative, with neutral element, 0X , and for every x ∈ X there is a y ∈ X such that x + y = 0X ; (2) (X, +X , ·X ) is a commutative ring. That is, ·X is also commutative, associative with neutral element, 1X (1X 6= 0X ), and is distributive over +X ; (3) There is an injection ϕX : N → X which is a homomorphism. That is for every n, m ∈ N

ϕX (n + m) = ϕX (n) +X ϕX (m), ϕX (n · m) = ϕX (n) ·X ϕX (m) and ϕX (1) = 1X ;

(4) (∀ y ∈ X)(∃ m, n ∈ N)(y + ϕX (n) = ϕX (m)). Note that there is no mentioning of any relation ≤ in this deﬁnition. In fact (see the exercise below), such a relation is uniquely determined if we want ϕX to be a monotone map, that is for m ≤ n ∈ N we have ϕX (m) ≤ ϕX (n).

Exercise 5.1.1 Suppose X is a set of integers, with corresponding map ϕX : N → X, and let x, y ∈ X. Define x ≤ y if either x = ϕX (n), y = ϕX (m) and n ≤ m, or x∈ / Ran(ϕX ) and y ∈ Ran(ϕX ), or x + ϕX (n) = 0, y + ϕX (m) = 0 and m ≤ n. Prove that ≤ defines a total order on X. A very important fact is that the set of integers is essentially unique. This means that every two such sets can be identified by a uniquely determined bijection. Next theorem gives the precise claim.

Theorem 5.1.1 Let X and X0 be two sets of integers, then there is a unique map ψ : X → X0 such that the triangle commutes (that is, ϕX0 = ψ ◦ ϕX )

∃!ψ X / X0 _ >

ϕX ϕX0 N In addition the map ψ is a homomorphism which is a bijection (so, ψ is an isomorphism of rings).

Proof (Sketch) (1) (Existence of ψ). Let x ∈ X and let x + ϕX (m) = ϕX (n) for some (non- 0 0 0 0 unique!) m, n ∈ N. There is a unique x ∈ X such that x + ϕX0 (m) = ϕX0 (n). Deﬁne ψ(x) = x . 0 This deﬁnition is correct, because if x + ϕX (m1) = ϕX (n1), then x + ϕX0 (m1) = ϕX0 (n1) as well. It is straightforward to check that for this map ψ one has ϕX0 = ψ ◦ ϕX . The himomorphism and bijection claims are left to the reader to prove. (2) (Uniqueness of ψ) Suppose ψ0 : X → X0 is 0 another such map. By the commutativity relation ϕX0 = ψ ◦ ϕX we get that if x ∈ Ran(ϕX ), then

0 0 ψ (x) = (ψ ◦ ϕX )(m) = ϕX0 (m) = (ψ ◦ ϕX )(m) = ψ(x)

0 and if x∈ / Ran(ϕX ), then x + ϕX (m) = ϕX (n), and since ψ is a homomorphism, we have on one hand 0 0 0 0 ψ (x + ϕX (m)) = ψ (x) + (ψ ◦ ϕX )(m) = ψ (x) + ϕX0 (m), 0 0 and on other ψ (x + ϕX (m)) = (ψ ◦ ϕX )(n) = ϕX0 (n). This gives us that

0 ψ(x) + ϕX0 (m) = ϕX0 (n) and ψ (x) + ϕX0 (m) = ϕX0 (n) which immediately implies that ψ0(x) = ψ(x). 2

68 Again - the theorem says that if integers exist, they are essentially unique. What remains to prove now is that a set of integers exists. This is what we are doing in the following sub-sections. The resulting set, as well as any other set of integers, will be denoted by Z.

Remark 5.1.1 The concepts of a (commutative) group and a (commutative) ring are examples of algebraic structures more about which one learns in the courses on Algebraic Structures and Topics in Algebraic Structures. Both, the group (Z, +), and the ring (Z, +, ·) play central roles in Algebra. 2

To solve the original, non-simpliﬁed, equation from the beginning of this sub-section, we will have to expand the integers to include all solutions to that equation (with the sole restriction m 6= 0). This way, we will get the set of rational numbers, which are a ﬁeld - an example of yet another algebraic structure.

The moral here is: the integers and the rational numbers are constructed for the needs of Arithmetic and Algebra.

5.1.2 Construction of a Set of Integers: Z The Set Z Consider the set X = N × N. Deﬁne the relation R ⊆ X × X by ((n, m), (n0, m0)) ∈ R if n + m0 = n0 + m.

Exercise 5.1.2 Prove that R is an equivalence relation on X.

Deﬁnition 5.1.2 The quotient set X/R = (N × N)/R is called the set of integers, and is denoted by Z.

Exercise 5.1.3 (1) Prove that if m ≤ n, so that n = m + s for some s ∈ N, then R[(n, m)] = R[(s, 0)]. Prove also that (∀s1, s2 ∈ N)((s1 6= s2) → R[(s1, 0)] 6= R[(s2, 0)]). (2) Prove that if n ≤ m, so that m = n + s for some s ∈ N, then R[(n, m)] = R[(0, s)]. Prove also that (∀s1, s2 ∈ N)((s1 6= s2) → R[(0, s1)] 6= R[(0, s2)]). (3) Conclude from (1) and (2) that

Z = {R[(s, 0)],R[(0, s)],R[(0, 0)] | 0 6= s ∈ N}, and that the RHS of the equality is a family of disjoint sets. In other words, every integer has a unique representative (n, m) where at least one of the components of the order pair is 0.

The Ordered Ring Z We introduce next two operation and a relation on Z which will make this set and ordered ring.

Deﬁnition 5.1.3 (1) The addition + : Z × Z → Z is deﬁned by R[(n, m)] + R[(n0, m0)] := R[(n + n0, m + m0)].

(2) The multiplication · : Z × Z → Z is deﬁned by R[(n, m)] · R[(n0, m0)] := R[(nn0 + mm0, nm0 + n0m)].

The relation ≤ on Z is deﬁned by R[(n, m)] ≤ R[(n0, m0)] if n + m0 ≤ n0 + m.

69 In this definition, there is a hidden problem which has to be dealt with. Namely, all the three items of the definition use particular representatives of the classes involved in order to define the RHSs in (1) and (2) or check the relation in (3)! The first thing to convince ourselves here is that no choices affect the results in each of the three items.

Proposition 5.1.2 The deﬁnitions of the two operations, + and ·, and of the relation ≤ above do not depend on any choices made.

Proof. (Sketch) For the addition, we have to show that if R[(n, m)] = R[(n1, m1)] and if 0 0 0 0 0 0 0 0 R[(n , m )] = R[(n1, m1)], then R[(n + n , m + m )] = R[(n1 + n1, m1 + m1)]. But the ﬁrst two 0 0 0 0 equalities mean that n + m1 = n1 + m and n + m1 = n1 + m which immediately implies that

0 0 0 0 0 0 (n + m1) + (n + m1) = (n1 + m) + (n1 + m ) = (n1 + n1) + (m + m ) due to the commutativity and associativity of addition in N. This proves our claim. For the multiplication, one has to show that

0 0 0 0 0 0 0 0 R[(nn + mm , nm + n m)] = R[(n1n1 + m1m1, n1m1 + n1m1)] which is left as an exercise. 0 0 0 0 For the relation, one has to show that if R[(n, m)] ≤ R[(n , m )], then R[(n1, m1)] ≤ R[(n1, m1)] 0 0 as well. We have, by the hypothesis, that n + m ≤ n + m, and that n + m1 = m + n1 and 0 0 0 0 0 0 n + m1 = m + n1. We want to show that n1 + m1 ≤ n1 + m1. To this end, observe that

0 0 0 0 0 0 n + m ≤ m + n ⇒ (n + m ) + (n1 + m1) ≤ (m + n ) + (n1 + m1)

0 0 0 0 0 0 0 0 ⇒ (n + m1) + (m + n1) ≤ (m + n ) + (n1 + m1) ⇒ (n1 + m) + (n + m1) ≤ (m + n ) + (n1 + m1) 0 0 0 0 0 0 ⇒ (m + n ) + (n1 + m1) ≤ (m + n ) + (n1 + m1) ⇒ n1 + m1 ≤ n1 + m1. All the implications are due to the properties of addition and ≤ in N. 2

Exercise 5.1.4 Verify that if

0 0 0 0 R[(n, m)] = R[(n1, m1)] and R[(n , m )] = R[(n1, m1)], then 0 0 0 0 0 0 0 0 R[(nn + mm , nm + n m)] = R[(n1n1 + m1m1, n1m1 + n1m1)]. The following theorem establishes some of the properties of the two operations, +, ·. According to these, (Z, +, ·) is a (commutative) ring with zero the class R[(0, 0)], and identity - the class R[(1, 0)].

Theorem 5.1.3 Let x, y, z ∈ Z. The following hold true for (Z, +, ·): (1) The addition is commutative, associative, has a neutral element, and every element has an opposite: x + y = y + x x + (y + z) = (x + y) + z (∀x)(x + R[(0, 0)] = x)(∀x)(∃x0)(x + x0 = R[(0, 0)]). (2) The multiplication is commutative, associative, and has a neutral element

x · y = y · x x · (y · z) = (x · y) · z

(∀x)(x · R[(1, 0)] = x). (3) The multiplication distributes over the addition

x · (y + z) = x · y + x · z.

70 Proof. See the exercise below. 2

Exercise 5.1.5 Prove the just stated theorem.

Exercise 5.1.6 (1) Prove that the neutral element of the addition is unique. That is

((∀w)(∀x)(x + w = x)) → (w = R[(0, 0)]).

This unique element is denoted by 0Z. (2) Prove that the neutral element of the multiplication is unique. That is

((∀w)(∀x)(x · w = x)) → (w = R[(1, 0)]).

This unique element is denoted by 1Z. (3) Prove that the opposite elements are unique. That is

(∀R[(n, m)])(∀x)((R[(n, m)] + x = 0Z) → (x = R[(m, n)])). The opposite element of x is denoted by −x. Prove also that

(∀x)(−(−x) = x).

(4) Prove that, for x, y ∈ Z we have (−x)y = x(−y) = −(xy) and (−x)(−y) = xy.

We are ready to solve the equation, for any a, b ∈ Z, X + a = b, as promised. The (unique!) solution is given by X0 = b + (−a).

The next theorem establishes the main properties of the relation ≤ including its relationship with the two operations on Z.

Theorem 5.1.4 The relation ≤ on Z has the following properties. (1) It is a total order, which has the least upper and the greatest lower bound properties; (2) (Additive Cancellation) (∀x, y, z ∈ Z)((x + z ≤ y + z) ⇔ (x ≤ y)); (3) (Monotonicity of Addition) ((a ≤ b) ∧ (c ≤ d)) → (a + c ≤ b + d), and if in addition (a < b) ∨ (c < d), then a + c < b + d;

(4) (0Z < a · b) ⇔ ((0Z < a) ∧ (0Z < b)) ∨ (a < 0Z) ∧ (b < 0Z)). (5) (Monotonicity of Multiplication) Let y < z. Prove that for every 0Z < x ∈ Z we have y · x < z · x. Prove also that if x < 0Z, then z · x < y · x. Proof. (1): Showing that the order is total. Let x = R[(m, n)] and y = R[(m0, n0)]. Since N is totally ordered, we have for the natural numbers m + n0 and m0 + n that

either m + n0 ≤ m0 + n or m0 + n ≤ m + n0.

In the ﬁrst case we have x ≤ y while in the second we have that y ≤ x. In any case, x and y are ≤- comparable. Showing that (Z, ≤) has the greatest lower bound property. Let A ⊆ Z be a bounded below non- empty subset. So, there is a x0 ∈ Z such that for every y ∈ A we have x0 ≤ y. Consider the set 0 A = {y + (−x0) | y ∈ A }. 0 This set is obviously bounded below by 0Z. Therefore, for every z ∈ A we have z = R[(m, n)] such that n ≤ m. As we know then there is an s ∈ N such that n + s = m, and therefore z = R[(s, 0)] as well. Now consider the set 00 0 A = {s ∈ N | R[(s, 0)] ∈ A }.

71 00 0 We have that ∅= 6 A ⊆ N, and has therefore a least element s0. Let z0 = R[(s0, 0)] ∈ A be the 0 0 corresponding element in A . By the deﬁnition of A , we have that z0 = x0 + y0 where y0 ∈ A. We will show that y0 is a greatest lower bound of A. Indeed, let y ∈ A, and let z = x0 + y be its 0 0 corresponding element in A . We have z0 ≤ z in A . So, x0 + y0 ≤ x0 + y in Z. By item (2) we get that y0 ≤ y. We just proved that y0 is a lower bound of A. Since y0 ∈ A, it is the greatest lower bound of A. Proving that Z has the least upper bound property is left as an exercise. (2) Let x = R[(m, n)], y = R[(m0, n0)] and z = R[(m00, n00)]. The direction (⇐) is straightforward, and is left as an easy exercise. Proving (⇒). We have

x + z = R[(m + m00, n + n00)] and y + z = R[(m0 + m00, n0 + n00)].

Therefore x + z ≤ y + z means that

m + m00 + n0 + n00 ≤ m0 + m00 + n + n00.

By the cancellation property for the addition in N, we have that m + n0 ≤ m0 + n which means that x = R[(m, n)] ≤ R[(m0, n0)] = y as claimed. (3) The inequalities a ≤ b and c ≤ d give that

a + c ≤ b + c and b + c ≤ b + d

which imply that a + c ≤ b + d. The second claim in (3) is done by, say, RAA, and is left as an exercise. (4) The direction (⇐) is straightforward. The opposite direction, (⇒) is proved by RAA, and is left as an exercise.

(5) This follows directly from item (4). Indeed, y < z means that 0Z < z + (−y), and by (4), since 0Z < x, we have 0Z < (z + (−y)) · x = zx + (−y)x = zx + (−(yx)). Therefore yx < (zx + (−(yx))) + yx = zx and yz < xz as claimed. The second claim in item (5) is left as an exercise. 2

Exercise 5.1.7 (1) Let A ⊆ Z, and B = {x | − x ∈ A }. Prove that A is bounded above if, and only if, B is bounded below. Conclude that if A is bounded above, then A has a least upper bound (which belongs to A). [Use that Z has the greatest lower bound property.] (2) Prove the second claim in item (3) of the theorem above. (3) Prove item (4) of the theorem above. (4) Prove the second claim in item (5) of the theorem above.

Remark 5.1.2 A ring with a total partial order (R, +, ·, ≤) is called an ordered ring if for every x, y, z ∈ R we have that (x < y → x + z < y + z) and ((x < y) ∧ (0 < z)) → (xz < yz). The theorem above conﬁrms in particular that (Z, +, ·, <) is an ordered ring. It can be proved that any ordered ring has a copy of (Z, +, ·, ≤) in itself. 2

Exercise 5.1.8 (1) Prove that if (a < 0Z)∧(b < 0Z), then a+b < 0Z. Similarly, if (0Z < a)∧(0Z < b), then 0Z < a + b. (2) Prove that for any 0 6= s ∈ N

0Z < R[(s, 0)] ∧ R[(0, s)] < 0Z. (3) Prove that ≤ is not a dense relation. (4) Deﬁne the function | • | : Z → Z by x if 0 ≤ x |x| = Z −x if x < 0Z

72 Prove that

(i) (∀x)(0Z ≤ |x| ∧ (|x| = 0Z → x = 0Z); (ii) (∀x, y)(|x · y| = |x| · |y|); (iii) (∀x, y)(|x + y| ≤ |x| + |y|).

Deﬁnition 5.1.4 The set Z+ := {x ∈ Z | 0Z < x } is called the set of positive integers. The set Z− := {x ∈ Z | x < 0Z } is called the set of negative integers. The set Z≥0 := {x ∈ Z | 0Z ≤ x } is called the set of non-negative integers.

In these notations, Z≥0 = {R[(s, 0)] | s ∈ N }, Z+ = {R[(s, 0)] | 0 6= s ∈ N }, and Z− = {R[(0, s)] | 0 6= s ∈ N }.

Natural Numbers as a Subset of the Integers

It is obvious by the construction of Z that N and Z are disjoint sets. It is convenient though to identify a subset of Z which has the same properties as N does, that is a copy of N in Z. This is done by designing a function from N to Z such that both operations and the relation on N are mapped to the corresponding ones in Z. Such a function is called a monotone homomorphism. More speciﬁcally, we have the following.

Theorem 5.1.5 Consider the function ϕZ : N → Z deﬁned by

(∀n)(ϕZ(n) = R[(n, 0)].

This function is a monotone homomorphism w.r.t. the operations +, · and the relation ≤:

0 0 0 0 ϕZ(n + n ) = ϕZ(n) + ϕZ(n ) ϕZ(n · n ) = ϕZ(n) · ϕZ(n )

0 0 n ≤ n ⇔ ϕZ(n) ≤ ϕZ(n ).

Also, ϕZ is an injection with Range(ϕZ) = Z≥0.

Proof. Straightforward and simple. Note that ϕZ(0) = 0Z and ϕZ(1) = 1Z. 2

Using ϕZ we identify N to Z≥0, and denote every element of the latter by the name of the element of the former it is identiﬁed with: n := R[(n, 0)]. According to the notations we accepted then −n := R[(0, n)], so that

Z = Z− t {0Z} t Z+ = {0, −1, 1, −2, 2,..., −n, n, . . . }, and we are in the realm of the notations we have been used to since pre-school.

The Operation Subtraction in Z Speaking of pre-school, we have been taught that there is a third operation on Z: subtraction! The definition of this is as follows − : Z × Z → Z (x, y) 7→ x − y := x + (−y). Notice that this operation is an ”ugly” one: it is neither commutative, nor associative! It is maybe worthwhile mentioning that every integer is a difference of two natural numbers. Indeed, if x = R[(n, m)], then x = n − m. Keep in mind though that this presentation is by no means unique! (Why?) Looking back in our construction of Z, we see that the definition of integers we used was ”modeled” on this, post-factum, presentation.

73 5.2 The Rational Numbers 5.2.1 Motivations; Deﬁnition of the Set of Rational Numbers Having deﬁned and constructed the set of integers, we turn in this section to solving the equation

aX + b = c

for integer numbers a, b and c. Obviously, this problem reduces to solving

aX = d (d = c − b).

We observe first that if a = 0, then solution exists only if d = 0, and if this is the case, every integer is a solution. So, the interesting case to consider is when a 6= 0. In this case, solution in integers may or may not exist. If we want to be able to always find a solution, then we have to, as we did in the case of the integers, to expand the integers to a set which has the properties of Z, so it should be a ring having Z as a sub-ring, and which contains the solutions to our equation for every a 6= 0 and d. As it was in the case of the integers, we want this expansion to be minimal: ideally, it should contain no more than the solutions to the equation above. We observe second that to solve the equation above, it is enough the expansion of Z to have the property that for every non-zero integer n there exists an element x of the expansion so that n·x = 1. Any such expansion of Z is called a set of rational numbers. Here is the expected definition.

Deﬁnition 5.2.1 Let (Y, +Y , ·Y ) be a commutative ring. It is called a set of rational numbers, and denoted by Q, if

(1) (∀ 0 6= y ∈ Y )(∃w ∈ Y )(y ·Y w = 1Y ). A commutative ring with this property is called a ﬁeld;

(2) There is an injection ϕY : Z → Y which is a ring homomorphism, that is,

ϕY (m + n) = ϕY (m) +Y ϕY (n) ϕY (m · n) = ϕY (m) ·Y ϕY (n) ϕY (1) = 1Y ;

(3) (∀x ∈ Y )(∃m, n ∈ Z)((n 6= 0) ∧ (ϕY (n) ·Y x = ϕY (m))). As it was in the case of the integers, there is no mentioning about a relation ≤ on Y , but it is automatically induced by ϕY . This will be explained in the following sub-section.

The ﬁrst thing to prove here is that the sets of rational numbers are canonically identiﬁable with each other. Here is the respective theorem.

Theorem 5.2.1 For every two sets of rational numbers Y and Y 0, there is a unique map ψ : Y → Y 0 such that ψ ◦ ϕ = ϕ0 . That is, the triangle below commutes. Y Y

∃!ψ Y / Y 0 _ >

ϕ ϕ Y Y 0 Z The map ψ is a bijective homomorphism of rings (so, it is an isomorphism of rings). Proof (Sketch) The proof is similar to the one for integers in the previous subsection. (1) (Exis-

tence of ψ) If y ∈ Y is such that y ∈ Ran(ϕY ), and y = ϕY (n), then deﬁne ψ(y) = ϕY 0 (n). If

y∈ / Ran(ϕY ), then let (the non-unique!) n, m ∈ Z such that n 6= 0, and y ·Y ϕY (n) = ϕY (m). There 0 0 0 0 is a unique y ∈ Y such that y ·Y 0 ϕY 0 (n) = ϕY 0 (m). Deﬁne in this case ψ(y) = y . Again, one proves that this deﬁnition is correct: does not depend on the choice of m and n 6= 0 above. Then, one also checks that this ψ is a bijection and a ring homomorphism. (2) (Uniqueness of ψ) Let ψ0 be another such map. We have to show that ψ0 = ψ. Let y ∈ Y , and let n 6= 0 and m be integers 0 so that y ·Y ϕY (n) = ϕY (m). Since ψ is a homomorphism, we have 0 0 0 0 0 ψ (y ·Y ϕY (n)) = ψ (ϕY (m)) ⇒ ψ (y) ·Y 0 ψ (ϕY (n) = ψ (ϕY (m))

74 0 ⇒ ψ (y) ·Y 0 ϕY 0 (n) = ϕY 0 (m) 0 which, together with ψ(y) ·Y 0 ϕY 0 (n) = ϕY 0 (m) implies that ψ (y) = ψ(y). Since this is true for every y ∈ Y , the two functions are equal. 2

The theorem says that if a set of rational number exists, then it is essentially unique. What we have to prove next is that a set of rational numbers does exist. We are doing this in the next sub-section.

5.2.2 Construction of a Set of Rational Numbers: Q The Set Q

Consider the set X = Z × Z+. Deﬁne a relation R on X by

((x, n), (y, m)) ∈ R if x · m = y · n.

Exercise 5.2.1 Prove that R is an equivalence relation on X.

Deﬁnition 5.2.2 The quotient set X/R = (Z × Z+)/R is called the set of rational numbers, and is denoted by Q.

Remark 5.2.1 As it was in the case of integers, every rational number has a special, uniquely determined representative. This representative is an ordered pair (x, n) ∈ Z × Z+ such that the integers x and n share no common factors, diﬀerent from ±1. To prove this, one needs certain knowledge from elementary number theory, and we omit it here. 2

The Ordered Field Q We are introducing next two operations, addition and multiplication, and a relation, ≤, on Q which will make the set an ordered ﬁeld.

Deﬁnition 5.2.3 (1) The addition + : Q × Q → Q is deﬁned by

R[(x, n)] + R[(y, m)] := R[(xm + yn, nm)].

(2) The multiplication · : Q × Q → Q is deﬁned by

R[(x, n)] · R[(y, m)] := R[(xy, nm)].

(3) The relation ≤ is deﬁned by

R[(x, n)] ≤ R[(y, m)] if xm ≤ yn.

Proposition 5.2.2 The deﬁnitions of the two operations, +, ·, and of the relation ≤ do not depend on any choices made to formulate them.

Proof. A straightforward exercise. 2

Exercise 5.2.2 Prove the Proposition above.

The following theorem collects some of the properties of the operations +, ·, and actually states that Q is a ﬁeld.

75 Theorem 5.2.3 Let x, y, z ∈ Q. The following holds true for (Q, +, ·) (1) The addition is commutative, associative, has a neutral element, and every element has an opposite: x + y = y + x x + (y + z) = (x + y) + z (∀x)(x + R[(0, 1)] = x)(∀x)(∃x0)(x + x0 = R[(0, 1)]). (2) The multiplication is commutative, associative, has a neutral element, and every non-zero rational number has a reciprocal:

x · y = y · x x · (y · z) = (x · y) · z

(∀x)(x · R[(1, 1)] = x)(∀x)((x 6= R[(0, 1)]) → (∃y)(x · y = R[(1, 1)])). (3) The multiplication distributes over the addition

x · (y + z) = x · y + x · z.

Proof. The proof is straightforward , and is left as an exercise. 2

Exercise 5.2.3 Prove the theorem above.

Remark 5.2.2 As stated already, the claims in the theorem above mean that (Q, +, ·) is a field. A field is a commutative ring such that every non-zero element has a reciprocal. The ring Z is not a field. 2

Exercise 5.2.4 (1) Prove that the neutral element of the addition is unique. That is

((∀w)(∀x)(x + w = x)) → (w = R[(0, 1)]).

This unique element is denoted by 0Q. (2) Prove that the neutral element of the multiplication is unique. That is

((∀w)(∀x)(x · w = x)) → (w = R[(1, 1)]).

This unique element is denoted by 1Q. (3) Prove that the opposite elements are unique. That is

(∀R[(x, n)])(∀w)((R[(x, n)] + w = 0Q) → (w = R[(−x, n)])). The opposite element of x is denoted by −x. Prove also that

(∀x)(−(−x) = x).

(4) Prove that the reciprocal elements are unique, when they exist. That is

(∀x 6= 0Q)(∀y, z)((x · y = 1Q ∧ x · z = 1Q) → (y = z)).

−1 The unique reciprocal element corresponding to x 6= 0Q is denoted by x . Prove also that

−1 −1 (∀x 6= 0Q)((x ) = x).

(5) Prove that for x, y ∈ Q we have (−x)y = x(−y) = −(xy) and (−x)(−y) = xy.

(6) Prove that for every non-zero x, y ∈ Q we have (xy)−1 = x−1y−1.

76 We are ready to solve the equation, even for any a 6= 0Q, d ∈ Q, aX = d,

−1 as promised. The (unique!) solution is given by X0 = d · a .

The next theorem establishes the main properties of the relation ≤ including its relationship with the two operations on Q.

Theorem 5.2.4 The relation ≤ on Q has the following properties. (1) It is a total order, which is dense; (2) (Additive Cancellation) (∀x, y, z ∈ Q)((x + z ≤ y + z) ⇔ (x ≤ y)); (3) (Monotonicity of Addition) ((a ≤ b) ∧ (c ≤ d)) → (a + c ≤ b + d), and if in addition (a < b) ∨ (c < d), then a + c < b + d;

(4) (0Q < a · b) ⇔ ((0Q < a) ∧ (0Q < b)) ∨ (a < 0Q) ∧ (b < 0Q)). (5) (Monotonicity of Multiplication) Let y < z, and if x be rational numbers. If 0Q < x, then y · x < z · x, and if x < 0Q, then z · x < y · x.

Proof. We are proving item (1) only. The rest are left as an exercise. To prove (1), let x, y ∈ Q such that x = R[(a, n)] and y = R[(b, m)] The numbers am and bn are integers, so either am ≤ bn or bn ≤ am. In the former case x ≤ y, in the latter case y ≤ x. The order ≤ on Q is total. The order is also dense. Indeed, let x < y be two rational numbers. By item (2) we get that x + x < x + y and x + y < y + y. Therefore x < (x + y)/2 < y, and we are done (notice that (x + y)/2 ∈ Q). 2

Exercise 5.2.5 Prove items (2), (3), (4), and (5) of the theorem above.

Remark 5.2.3 A field with a total partial order (L, +, ·, ≤) is called an ordered field if for every x, y, z ∈ R we have that x < y → x + z < y + z and x < y ∧ 0 < z → xz < yz. The theorem above confirms in particular that (Q, +, ·, ≤) ordered field. It can be shown that every ordered field has a copy of (Q, +, ·, ≤) in itself. 2

Exercise 5.2.6 (1) Prove that if (a < 0Q)∧(b < 0Q), then a+b < 0Q. Similarly, if (0Q < a)∧(0Q < b), then 0Q < a + b. (2) Prove that

0Q < R[(x, n)] ⇔ x ∈ Z+. (3) Prove that (Q, ≤) has the least upper bound property if, and only if, it has the greatest lower bound property. (4) Deﬁne the function | • | : Q → Q by

x if 0 ≤ x |x| = Q −x if x < 0Q Prove that

(i) (∀x)(0Q ≤ |x| ∧ (|x| = 0Q → x = 0Q)); (ii) (∀x, y)(|x · y| = |x| · |y|); (iii) (∀x, y)(|x + y| ≤ |x| + |y|).

The Set Z as a Subset of Q It is obvious by the construction of Q that Z and Q are disjoint sets. It is convenient though to identify a subset of Q which has the same properties as Z does, that is a copy of Z in Q. This is done by designing a function from Z to Q such that both operations and the relation on Z are mapped to the corresponding ones in Q. Such a function is called a monotone homomorphism. More speciﬁcally, we have the following.

77 Theorem 5.2.5 Consider the function ϕQ : Z → Q deﬁned by

(∀x)(ϕQ(x) = R[(x, 1)]. This function is a monotone homomorphism w.r.t. the operations +, · and the relation ≤: 0 0 0 0 ϕQ(x + x ) = ϕQ(x) + ϕQ(x ) ϕQ(x · x ) = ϕQ(x) · ϕQ(x ) 0 0 x ≤ x ⇔ ϕQ(x) ≤ ϕQ(x ).

Also, ϕQ is an injection with Ran(ϕQ) = {R[(x, 1)] | x ∈ Z }..

Proof. Straightforward and simple. Note that ϕQ(0) = 0Q and ϕQ(1) = 1Q. 2 Exercise 5.2.7 Prove the theorem above.

Remark 5.2.4 Identifying Z with Ran(ϕQ) makes the former a subring of the ﬁeld Q. More about these - in the Algebra course. 2

Exercise 5.2.8 Show that for every integers x ∈ Z, ϕQ(|x|) = |ϕQ(x)|.

Using ϕQ we identify Z to {R[(x, 1)] | x ∈ Z }, and denote every element of the latter by the name of the element of the former it is identiﬁed with: x := R[(x, 1)]. According to the notations we accepted then −x := R[(−x, 1)]. On the other hand, every rational number R[(x, n)] is a product of two numbers from {R[(x, 1)] | x ∈ Z }: R[(x, n)] = R[(x, 1)] · R[(n, 1)]−1. Recalling that Z = {0, 1, −1, 2, −2, . . . , n, −n, . . . }, we arrive at the realm of the notations of rational fractions we have been used to since pre-school: −1 −1 Q = {(m) · n , (−m) · n | m ∈ Z≥0, n ∈ Z+} = {m/n, (−m)/n | m ∈ Z≥0, n ∈ Z+}.

Exercise 5.2.9 (Archimedes Property of Q) Prove that for any x ∈ Q, there is a y ∈ Z+ such that x < y.

[Hint: If x = m/n for m ∈ Z, let y = |m| + 1.]

Note that many rational fractions correspond to a rational number. As remarked in the beginning of this sub-section, every rational number R[(x, n)] has a unique representative: (x0, n0) ∈ R[(x, n)] determined by x0 and n0 not sharing common factors diﬀerent from ±1. The rational fraction x0/n0 represents that special element, and is called reduced fraction. So, every rational number has a unique reduced fraction representing it. In the new notations, m m m n + m n m m m m 1 + 2 = 1 2 2 1 1 · 2 = 1 2 . n1 n2 n1n2 n1 n2 n1n2

The operations Subtraction and Division in Q Speaking of pre-school, we have been taught that there is two more operations on Q: subtraction, and division! The deﬁnitions of these are as follows × − : Q × Q → Q ÷ : Q × Q → Q (x, y) 7→ x − y := x + (−y)(x, y) 7→ x ÷ y := x · y−1 × Here Q = Q \{0Q} is the set of non-zero rational numbers. Notice that the additional operations are ”ugly”: they are neither commutative, nor associative!

Similarly to what we had for integers, every rational number can be presented as a quotient of an integer and a natural number: x = m/n = m ÷ n. This presentation is not unique. Looking back in our deﬁnition of rational numbers we see that it was modeled on this, post-factum, presentation.

78 (Q, ≤) is not Continuous We end up the subsection on rational numbers proving that we have to work more in order to get a ﬁeld with continuous order on it.

Example-Proposition 5.2.3 The relation ≤ on Q is not continuous. More precisely, it doesn’t have the least upper bound property.

Proof It’s enough to exhibit a bounded above non-empty subset of Q which has no l.u.b. To this end, let X = {x ∈ Q | 0 < x ∧ x2 ≤ 2 }. This set is non-empty and is bounded above: 1 ∈ X, and (∀x)(x ∈ X → x < 2). We are showing next that X has no l.u.b. in Q. (1) We claim ﬁrst that if y ∈ Q is an upper bound of X, then 2 < y2. By way of contradiction, assume that y2 ≤ 2. Since, as we know, there are no rational numbers z such that z2 = 2 (Euclid’s theorem), then y2 < 2. Obviously, we have y = m/n for some positive integers m, n. Also, 0 < 2 − y2 ∈ Q, so 2 − (m/n)2 = a/b for positive integers a and b. Observe that, for every positive integer s we have

m2 sm + 12 m 1 2 m2 m 1 1 2 y2 = < = + = + 2 · + n sn n sn n n sn sn

a m 1 1 2 = 2 − + 2 · + b n sn sn the third equality, because (m/n)2 = 2 − a/b. Since (1/sn)2 ≤ 1/sn, and that 1/sn ≤ 1/s, we have that the last expression is a m 1 1 2m + 1 a ≤ 2 − + 2 · + = 2 + − . b n s sn sn b When s = b(2m + 1), we get that 2m + 1 a 1 an 2 + − = 2 + − ≤ 2. sn b bn bn

Combining all inequalities and equalities above, we arrive at (for s = b(2m+1) and a/b = 2−(m/n)2)

m2 sm + 12 y2 = < ≤ 2. n sn This result implies in particular that, for the chosen s, the rational number z = (sm + 1)/(sn) belongs to X. It also imlies that y < z. This latter is impossible, because y, an upper bound of X, is actually smaller than that rational number - a contradiction! Therefore, every upper bound y of X satisﬁes 2 < y2. (2) We claim now that there is no upper bound y ∈ Q of X which has 2 < y2. The argument is similar to the one in (1): for every upper bound y = m/n we will ﬁnd, for appropriate s ∈ N, another upper bound in the form y0 = (sm − 1)/(sn) ∈ Q of X. Note that in this case 2 > y2, and so 2 − (m/n)2 = c/d is a positive rational number. We have in these notations m2 ms − 12 m 1 2 m2 1 m 1 m2 2m > = − = − 2 − > − , n ns n ns n ns n ns n n2s and so, ms − 12 m2 2m c 2m − 2 > − 2 − = − . ns n n2s d n2s Choosing s = 2md, we get that c 2m c 1 − = − ≥ 0. d n2s d n2d

79 Therefore, for the chosen s, we have

ms − 12 2 < < y2 ns

as promised. 2

Vista: Solving All Algebraic Equations Our motivation for constructing the rational numbers came from the need for being able to solve all linear equations of one unknown with integer coefficients. In our construction, we added all solutions to such equations. The resulting extension naturally turned out to be a field. And indeed, Q is up to an isomorphism of fields, the smallest field where we can solve all linear equations with integer coefficients. If we want to be methodical in our study of number systems, we have to try to solve quadratic equations, cubic equations, and in general, ant degree equations with integer coefficients as well. This can certainly be done, and the methods for doing that are purely algebraic (more about this in the course of Galois Theory). The resulting extension of the integer numbers, naturally containing Q, is again a field, called the field of algebraic numbers, and denoted by A. This field is, up to isomorphism, the smallest field containing the solutions to all polynomial equations with integer coefficients (n ≥ 1)

n n−1 f(X) = anX + an−1X + ··· + a1X + a0 where an, . . . , a0 ∈ Z. What is more significant about the field A is that it is algebraically closed, that is it consists of the solutions to all polynomial equations with coefficients in itself. In other words, there is no need to extend A if we want to solve polynomial equations with coefficients in A.

5.3 The Real Numbers

In this section we construct the important for Calculus, and for the whole math!, set of real numbers. The construction is based on the notion of Dedekind cut. An Alternative deﬁnition is presented later on in Chapter 8 as well. The rest of the course will be devoted to studying this set.

5.3.1 Motivations; Deﬁnition of the Set of Real Numbers Before giving a deﬁnition of real numbers, we discuss what we want properties this set is desirable to have.

As we know the set of rational numbers Q is an ordered field containing the integers Z as an ordered subring. This field Q is better than the ring Z, because we can solve linear equations in it. For this we didn’t need the order on Q, and as we know the order on both Z and Q is determined by the fact that they have the natural numbers N as a subset. The order on Q unfortunately is not convenient for doing Calculus where limits of sequences are important: the order doesn’t have the least upper bound property, and so, although being dense, is not continuous). The goal in constructing real numbers therefore is to extend the rational numbers in such a way that we keep the properties the latter have (related to addition, multiplication, and the order), but to have the order continuous. So, the real numbers should be an ordered field containing the rationals as an ordered sub-field, and having the order continuous. We would also like to have, as it was in the case of Z extending N, and of Q extending Z, the real numbers as economically defined as possible. Recall that in the case of integers we used the addition as a means to define minimality:

(∀m ∈ Z)(∃s ∈ N)(m + s ∈ N). In the case of rational numbers, we used the other operation to deﬁne the minimality:

(∀x ∈ Q)(∃n ∈ Z)(0 < n ∧ n · x ∈ Z).

80 Both these minimality conditions were based on algebra (solving equations). It is natural therefore to deﬁne the minimality of the real numbers using the relation. The way to assure this minimality here is by demanding to have the rational numbers a dense subset of the real numbers.

Our discussion motivates the following deﬁnition of what real numbers are.

Definition 5.3.1 An ordered filed (K, +, ·, ≤) with continuous order is called a set/field of real numbers if there is a monotone homomorphism ϕK : Q → K such that Ran(ϕK ) ⊆ K is a dense subset.

Recall that monotone homomorphism means that for every x, y ∈ Q we have

ϕK (x + y) = ϕK (x) + ϕK (y) ϕK (x · y) = ϕK (x) · ϕK (y) ϕK (1) = 1K and

x ≤ y ⇔ ϕK (x) ≤ ϕK (y). There is a bit of terminology which makes this deﬁnition more structured. Here it is.

Definition 5.3.2 (1) An ordered field (K, +, ·, ≤) with a continuous order ≤ is called a complete ordered field. (2) A monotone homomorphism f : Q → K where K is a complete ordered field is called an ordered completion of Q if Ran(f) ⊆ K is a dense subset. Therefore a set of real numbers is any ordered completion of Q.

Remark 5.3.1 Recall that a ﬁeld L is ordered if there is a total partial order ≤ on it which respects the operations of the ﬁeld. That is, for every x, y, z ∈ L we have

x ≤ y → x + z ≤ y + z and x ≤ y ∧ 0 ≤ z → x · z ≤ y · z.

It is not hard to conclude from here that 0 ≤ 1, and therefore that 0 < 1. With a little bit more work one shows then that there is a unique ring homomorphism ϕL : Q → L, and that in addition this homomorphism is monotone (and so - injective). If furthermore the relation ≤ is continuous, then Ran(ϕL) is dense in L. So, any complete ordered field contains a unique copy of the rational numbers as a subring. If there is no harm caused, this unique copy is identified with Q. Therefore, any complete ordered field L which has Q as a dense subset is a completion of Q. 2 The potential ambiguity caused by the non-uniqueness of the set of real numbers is taken care of by the minimality property of this set: the real numbers are essentially unique. Here is the corresponding theorem.

Theorem 5.3.1 Suppose ϕK1 : Q → K1 and ϕK2 : Q → K2 are two ordered completions of Q.

Then there is a unique map ψ : K1 → K2 such that ψ ◦ ϕK1 = ϕK2 . Moreover, this map is a bijection, and is a monotone homomorphism (that is, ψ is an isomorphism of ordered ﬁelds). Proof The proof will be given in the end the this section. 2

By using this unique map, we an identify the fields K1 and K2 not only as sets, but, due to the map being an isomorphism of ordered fields, as ordered fields as well. We will construct in the following section a set of real numbers using Dedekind cuts. Later in the course, we will construct a set of real numbers based on Cauchy sequences of rational numbers. The two sets are, of course, different, but, due to the theorem above, can be canonically/unambiguously identified.

Speaking of the concrete construction of real numbers, motivations for that come both, from algebra and from geometry. The former is related to formalizing the notion of limits of sequences, and ensure they exist. That can be done purely algebraically, by using a method suggested by Can- tor, among other mathematicians of his time, and is based on the algebra of sequences of rational

81 numbers. We will discuss this method when we introduce rigorously the basics of sequences and their limits. Turning point in this approach is the notion of Cauchy sequences: the ones that we want to have limits. The geometric approach follows the lead of ancient Greek mathematicians in their attempt to ensure non-existence of gaps on the lines in the Euclidean geometry. More about all this can be learned in the course on College Geometry, but in a nutshell it can be described as follows. The Greeks were interested in positive numbers only. The reason for this was purely geometric. Any two points on a given line define a segment. The segment has a length - a positive number. Numbers for Greeks were synonymous with lengths of segments. Having chosen a segment of unit length, the Greeks were able to construct segments of length a (positive) rational number, and these segments corresponded to points on a ray. For some time, the Greeks thought that no more points existed on that ray. Equivalently, they√ thought that every segment has a length expressible as a rational number. But after proving that 2, the length of the diagonal of a square of side lengths 1, was not a rational number, they realized that was not the case... Our (first) construction of the set of real numbers, based on the theory of Dedekind cuts, follows the steps of the great Greek mathematicians: we will define numbers corresponding to every point on that ray.

5.3.2 Construction of a Set of Real Numbers: R Deﬁnition 5.3.3 A Dedekind cut is an ordered pair (α, β) of subsets of Q such that (1) {α, β} is a partition of Q, that is

α ∪ β = Q α 6= ∅= 6 β α ∩ β = ∅; (2) (∀a ∈ α)(∀b ∈ β)(a < b); (3) α has no maximum element in Q. In other words,

¬((∃x ∈ Q)(α = {y ∈ Q | y ≤ x })).

For every Dedekind cut (α, β), the set α is an initial segment in Q, and the set β = Q \ α is the relative complement of α. It is straightforward that β is a ﬁnal cut of (Q, ≤). Obviously, the Dedekind cut (α, β) is uniquely determined by either α or β. In what follows, we will identify the cut (α, β) with the initial segment α. In these terms, a Dedekind cut is a non-empty initial segment in Q which is a proper subset of Q, and has no maximal element.

We deﬁne the set that we will prove constitutes a set of real numbers.

Deﬁnition 5.3.4 The set R of real numbers is deﬁned by

R := {α | α is a Dedekind cut }. The set β of a pair (α, β) has the property that

(∀x)((x ∈ β) → (∀z ∈ Q)((x ≤ z) → z ∈ β)),

and it is natural to call β a ﬁnal segment in Q.

There are Dedekind cuts determined by rational numbers.

Deﬁnition 5.3.5 Let x ∈ Q. The set {z ∈ Q | z < x} is called a rational cut, and is denoted by x∗.

Exercise 5.3.1 (1) Prove that a rational cut is a Dedekind cut. (2) Prove that (∀x, y ∈ Q)((x 6= y) → (x∗ 6= y∗)). (3) Show that for every x ∈ Q, the set x∗ has a least upper bound (in Q) - the rational number x itself. Moreover, prove that a Dedekind cut α is rational if, and only if, α has a l.u.b. in Q.

82 The Exercise teaches us that the rational cuts are characterized as those Dedekind cuts who have l.u.b. in Q. A natural question here is: are there non-rational (called irrational) cuts? The answer is Yes as the following exercise suggests to prove.

Exercise 5.3.2 Prove that 2 α := {y ∈ Q | (y ≤ 0) ∨ (y ≤ 2)} is a non-rational Dedekind cut.

5.3.3 The Relation ≤ on R Deﬁnition 5.3.6 Let α, β ∈ R. we say that α is less than or equal to β, and write α ≤ β, if α ⊆ β.

The following theorem says that (R, ≤) is a continuous relation! But before that - some of the properties of families of initial segments in Q in exercises.

Exercise 5.3.3 Let A := {αλ | λ ∈ Λ } be a non-empty family of initial segments in Q. Prove that (1) If ∩A= 6 ∅, then ∩A is an initial segment in Q; (2) If ∪A= 6 Q, then ∪A is a bounded above initial segment of Q.

Theorem 5.3.2 The relation ≤ is a total order on R, it is dense, and has the least upper bound property.

Proof. That the relation ≤ is a partial order is immediate (an exercise to do below). We are verifying it’s connected: given α, β ∈ R, either α ≤ β or β ≤ α. To this end, suppose ¬(α ≤ β). This means that α * β which happens only when there is a rational number x ∈ α \ β. Since β is an initial segment of Q, for every y ∈ β we have y < x. For the reason that α is an initial segment in Q now, all elements y ∈ β, being less than x ∈ α, belong to α. Therefore β ⊆ α, and β ≤ α. That the relation ≤ is dense follows from the fact that the rational cuts are dense in R - an exercise to do below. We are showing that the relation ≤ has the least upper bound property. Let X ⊆ R be a non-empty bounded above subset. The elements of X are Dedekind cuts, so X = {αx | x ∈ X}. The set X is bounded above: there is y ∈ R such that for every x ∈ X we have x ≤ y. And, of course, y = αy is a Dedekind cut. Consider the set

z = ∪X = ∪{αx | x ∈ X} ⊆ Q.

Obviously z ⊆ αy, it is an initial segment in Q as a union of such, and has no maximum element, because the sets αx don’t have maximum elements (ﬁll in the details as an exercise). This means that z = β is a Dedekind cut. Since, for every x ∈ X, we have αx ⊆ β, the real number z is an upper bound of X. We are showing now that z is the least upper bound of X. Indeed, if that’s not the case, there should exist an upper bound w ∈ R of X such that w < z. But this leads to

∪X ⊆ w $ z = ∪X

which in turn leads to ∪X $ ∪X - a contradiction. The theorem is proved. 2

The following exercise is a prelude to deﬁning the monotone homomorphism ϕQ : Q → R needed to show that R is a set of real numbers.

Exercise 5.3.4 (1) Prove that (R, ≤) is a partial order. (2) Prove that the rational cuts are dense in R as well. That is, for every two distinct real numbers α and β, there is a rational number x ∈ Q such that α < x∗ < β ∨ β < x∗ < α.

83 (3) Prove that the set z in the proof above is a Dedekind cut. (4) (Archimedes property for R) Prove that for every Dedekind cut α, there is a natural number n such that α < n∗.

Deﬁnition 5.3.7 The positive real numbers are deﬁned by

∗ R+ := {α ∈ R | 0 < α }. The negative real numbers are deﬁned by

∗ R− := {α | α < 0 }.

∗ So, we have R = R− t {0 } t R+.

The set R contains a copy of the rational numbers, and has a continuous relation on it. We have to deﬁne next two operations on R, extending the addition and the multiplication on Q, and show that they have properties similar to their counterparts in Q. This is harder than anything we have done in this course so far. We do that in the next subsection.

5.3.4 The Operations Addition and Multiplication in R Before we introduce the operation addition on R, let’s ”play” a bit with real numbers.

Exercise-Proposition 5.3.5 Let α and β be real numbers. Prove that the set

X := {y ∈ Q | (∃z ∈ α)(∃w ∈ β)(y = z + w) } is a bounded above initial segment in Q which has no maximum element. So, X = γ ∈ R.

Exercise-Proposition 5.3.6 Let α be an initial segment in Q, and let α := {x ∈ Q | − x ∈ α }. Denote the relative complement of α in Q by Y := Q \ α. Prove that Y is an initial segment as well. In particular, if Y has no maximum element, then it s a real number.

Deﬁnition 5.3.8 Let α ∈ R, and let Y := Q \ α. Deﬁne Y if Y has no maximum element −α = Y \{m} if m is the maximum element of Y

Exercise 5.3.5 (1) Let α ∈ R, deﬁne Z := {x ∈ Q | (∃y ∈ α)(∃z ∈ −α)(x = y + z) }. Prove that Z = 0∗. [Hint: Notice that z ∈ aα implies that −z∈ / α, and that therefore it is an upper bound for α. If in addition y ∈ α, we have y < −z which implies y +z < 0, and therefore y +z ∈ 0∗. This proves Z ⊆ 0∗. Conversely, if x < 0, there is a y ∈ α such that y + (−x) ∈/ α (here, one may have to use the Archimedes property of Q). Therefore, −(y + (−x)) ∈/ α, and so, z := x − y ∈ −α. This shows that x = y + z where y ∈ α and z ∈ −α. We get that 0∗ ⊆ Z.] (2) Prove that −(−α) = α, and that α ∈ R+ ⇔ −α ∈ R−.

These exercises, and the Propositions above motivate the following deﬁnition.

Deﬁnition 5.3.9 The operation addition + : R × R → R is deﬁned by

(α, β) 7→ α + β := {y ∈ Q | (∃z ∈ α)(∃w ∈ β)(y = z + w)}.

And indeed, we have the following theorem.

84 Theorem 5.3.3 The operation addition on R is commutative, associative, with a (unique!) neutral ∗ element, 0R = 0 , and with (unique!) opposite elements (the opposite of α is = −α.) Proof In class and as an exercise. 2

Moreover, the interaction of the addition with the relation ≤ is also nice.

Theorem 5.3.4 For every four real numbers α, β, γ, and δ we have

α ≤ β ⇔ α + γ ≤ β + γ;

(α ≤ β) ∧ (γ ≤ δ) → (α + γ ≤ β + δ) where if there is a strict inequality anywhere on the LHSs above, then the inequality on the corresponding RHS is strict as well.

Proof. In class and as an exercise. 2

We turn now to the deﬁnition of multiplication on R. As it was in the case of addition, we need to prepare our ground for the corresponding deﬁnition.

Exercise-Proposition 5.3.7 Let α, β ∈ R+, Consider the sets

X(α,β) := {x ∈ Q | (∃y∈ / α)(∃z∈ / β)(x = y · z) }. and Y(α,β) := Q \ X(α,β). Then ∅= 6 Y(α,β) ⊂ Q, and is an initial segment in Q. Proof. In class and as an exercise. 2

Let’s deﬁne, for positive real numbers only!, the following operation

Deﬁnition 5.3.10 Let α, β ∈ R+. Deﬁne, in the notations of the Proposition-Exercise above,

Y if Y has no maximum element α β = (α,β) (α,β) Y(α,β) \{m} if m is the maximum element of Y(α,β)

Again for positive real numbers only we do the following.

Exercise-Proposition 5.3.8 Let α ∈ R+. Consider the family of positive rational cuts

∗ ∗ −1 ∗ A := {x | (x ∈ Q) ∧ (0 < (x ) < α) }.

Then ∩A is a non-empty bounded initial segment in Q containing a positive rational number. Proof. In class and as an exercise. 2

Deﬁnition 5.3.11 Let α ∈ R+. Deﬁne, in the notations of the previous Proposition-Exercise,

∩A if ∩ A has no maximum element α˜ = ∩A \ {m} if m is the maximum element of ∩ A

Note that, for α ∈ R+, theα ˜ is a positive real number.

∗ Exercise 5.3.6 Prove that (∀α ∈ R+)(α α˜ = 1 ). We have enough reasons now to make the following deﬁnition.

85 Deﬁnition 5.3.12 The multiplication · : R × R → R is deﬁned by   α β if α, β ∈ R  −((−α) β) if α ∈ , β ∈ (α, β) 7→ α · β = R− R+  (−α) (−β) if α, β ∈ R−  0∗ if α = 0∗ ∨ β = 0∗

The anticipated theorem here is the following one.

Theorem 5.3.5 The multiplication · on R is commutative, associative, has a(unique) neutral ele- ∗ ment, 1R = 1 , distributes over the addition, and every non-zero real number, α, has a unique (!) reciprocal, α˜ if α ∈ α−1 = R+ −(−fα) if α ∈ R− Proof. In class, and a slightly harder exercise. 2

We know now that R is a ﬁeld, and that the relation ≤ is continuous. To have the picture complete, we need to know the way the multiplication interacts with the relation. We have the following result.

Theorem 5.3.6 Let α, β ∈ R, γ ∈ R+, and δ ∈ R−. We have

(α ≤ β) → (α · γ ≤ β · γ) ∧ (β · δ ≤ α · δ).

Proof. In class, and an exercise. 2

The Set Q as a subset of R As it was with the case of integers and rationals, the set R has nothing to do with the set Q. But, again as it was before, we can construct a function with domain Q which locates an ”exact” copy of Q in R. To this end, consider the map

ϕR : Q → R

∗ x ∈ Q 7→ ϕR(x) := x .

Theorem 5.3.7 The map ϕR : Q → R is a monotone homomorphism of ordered ﬁelds. That is

ϕR(x + y) = ϕR(x) + ϕR(y), ϕR(x · y) = ϕR(x) · ϕR(y)

ϕR(1) = 1R, x ≤ y ⇔ ϕR(x) ≤ ϕR(y). Proof. An easy one even for an exercise! 2

The needed copy of Q in R is the the range Ran(ϕR), the rational cuts in R. We may, and will, assume from now on that N ⊂ Z ⊂ Q ⊂ R. Recall that this copy of Q is dense in R.

Exercise 5.3.7 (1) Deﬁne the absolute value function| • | : R → R the usual way, and prove that, for every x ∈ Q,

|ϕR(x)| = ϕR(|x|). (2) Prove that R has the Archimedean property that

(∀x ∈ R)(∃n ∈ N)(x < n).

86 The Operations Subtraction and Division on R Following the tradition, we end this subsection by mentioning the existence of two more operations × on R: subtraction and division. Denoting R = R− ∪ R+, we have

− : R × R → R (α, β) 7→ α − β := α + (−β) and × −1 ÷ : R × R → R (α, β) 7→ α ÷ β :=: α/β := α · β . Again respecting the tradition we emphasize that these two operations are ”ugly”: they are neither commutative, nor associative.

Completions of Q, and Uniqueness of the Real Numbers As it was the case of integers and rationals, the real numbers are essentially unique. Here is what that means.

Theorem 5.3.8 Let K be a ﬁeld with continuous order ≤K on it, and such that there is a monotone homomorphism ψ : Q → K with Ran(ψ) dense in K. Then, there is a unique monotone map FK : K → R such that FK ◦ ψ = ϕR.

∃!F K K / _ ∼= ? R

ϕ ψ R Q

The homomorphism FK is a ﬁeld homomorphism which is a bijection. This unique monotone map is used to identify K and R in a unique way. We say in this case that K and R are canonically identiﬁable. In this sense is R essentially unique. Before we give a sketch of a proof, recall a useful terminology.

Definition 5.3.13 The ordered field (K, +, · ≤) is called a completion of Q, if (1) (K, ≤) is a continuous order, and (2) There is a monotone homomorphism ψ : Q → K with Ran(ψ) ⊆ K a ≤-dense subset. The theorem above states that the completion of Q is essentially unique. Later on in the course, we will give a sketch of the construction of a completion RC of Q invented by Cantor (the details occupy Chapter 8 of the complete LN). Due to this theorem, the two completions, R and RC, are canonically identifiable.

Proof of Theorem (Sketch) The idea of constructing FK is simple and natural. First off, observe that the relation ϕR = FK ◦ ψ determines uniquely the map FK on Ran(ψ) ⊆ K. On the other hand, the elements of K define Dedekind cuts as follows: for x ∈ K define

Dx = {y ∈ Q | ψ(y) < x } ⊆ Q.

But any Dedekind cut IS a real number, so we deﬁne FK (x) = Dx ∈ R. It is straightforward that if x = ψ(q) ∈ Ran(ψ), then

Dx = Dψ(q) = ϕR(q), so that with this deﬁnition of FK we have ϕR = FK ◦ ψ. A direct check shows that FK is monotone. This proves the existence of FK .

The uniqueness of FK follows from the relation ϕR = FK ◦ ψ, and the restriction on FK to be monotone. Indeed, under these conditions, the map FK should map the subset

D˜ x = Ran(ψ) ∩ {y ∈ K | y < x}

87 −1 ˜ bijectively to ϕR(ψ (Dx)) = ϕR(Dx). The map FK is one-to-one, because it is monotone. It is onto, because (K, ≤) is continuous: every Dedekind cut of Q defines, through ψ, a bounded above initial segment of (Ran(ψ), ≤), and therefore, through the least upper bound of this segment, an element x ∈ K which is mapped by FK to the real number determined by that Dedekind cut. Finally, FK is a homomorphism of fields for the reason that the subsets D˜ x ⊆ K play the role of Dedekind cuts of Ran(ψ), and that the operations addition and multiplication on K can be identified with the respective operations on Dedekind cuts the same way it was done for R. 2

We are ready to prove now the theorem formulated in the beginning of this section.

Theorem 5.3.9 Suppose ϕK1 : Q → K1 and ϕK2 : Q → K2 are two completions of Q. then there

is a unique map ψ : K1 → K2 such that ψ ◦ ϕK1 = ϕK2 . Moreover, this map is a bijection, and is a monotone homomorphism (that is, ψ is an isomorphism of ordered ﬁelds). Proof The argument is based on the result from the previous theorem. Existence of ψ. By the

theorem above, there are unique maps FKi : Ki → R such that ϕR = FKi ◦ ϕKi for i = 1, 2. We know that F , i = 1, 2 are monotone isomorphisms as well. Let ψ = F −1 ◦ F : K → K . It is Ki K2 K1 1 2 straightforward that ψ is a monotone isomorphism (as a composition of such!). We also have ψ ◦ ϕ = (F −1 ◦ F ) ◦ ϕ = F −1 ◦ (F ◦ ϕ ) K1 K2 K1 K1 K2 K1 K1 = F −1 ◦ ϕ = F −1 ◦ (F ◦ ϕ ) = ϕ , K2 R K2 K2 K2 K2 so that ψ satisﬁes the condition from the theorem. Uniqueness of ψ. Suppose there is another 0 0 (set) map ψ : K1 → K2 such that ψ ◦ ϕK1 = ϕK2 . We have 0 0 (FK2 ◦ ψ ) ◦ ϕK1 = FK2 ◦ (ψ ◦ ϕK1 ) = FK2 ◦ ϕK2 = ϕR = FK1 ◦ ϕK1 , and so 0 (FK2 ◦ ψ ) ◦ ϕK1 = ϕR and FK1 ◦ ϕK1 = ϕR. But this means that the maps F ◦ ψ0 and F have to be the same, and so, ψ0 = F −1 ◦ F = ψ. K2 K1 K2 K1 The uniqueness is proved. 2

Remark 5.3.2 An ordered ﬁeld (L, +, ·, ≤) contains, as mentioned already, a copy of (Q, +, ·, ≤) in itself. If in addition the copy of Q is dense in L, and if ≤ is continuous, then L is a completion of Q. The theorem above gives than that this L is canonically isomorphic with R. 2 Having constructed the real numbers R, we are ready to begin our introduction to Advanced Cal- culus.

Vista: Other Completions of Q. In this section we constructed the only, up to isomorphism, ordered completion of the field of rational numbers. The starting point was the total order on Q induced by the order on N. The goal was to find an extension of Q where ”limits of sequences exist”. More precisely, as we will see in the next Chapter, we want every Cauchy sequence to have a limit. Cauchy sequences are defined using the absolute value function on Q (extended to R). Turns out, every function on Q like the absolute value one, determines corresponding Cauchy sequences (of rational numbers). Moreover, a method invented by Cantor produces an extension of Q in which every Cauchy sequence has a limit (see Chapter 8 of the complete LNs). The amazing fact here is that all absolute value-like functions, with the exception of the one that we used here, have number theoretical nature! The list of such functions contains essentially the absolute value defined in this Chapter, and the p-adic absolute values for p a prime number (a Theorem by Alexander Ostrowski). Those completions are denoted by Qp, and are called p-adic numbers. They are not ordered, of course! The set R is one of denumerably many completions of Q, and it is the only transcendental, that is, non-algebraic, one. Essentially, all methods developed to study R and the functions on it have been extended to study Qp. A bit more about all this can be learned in the course MAS 4203 Number Theory.

88 Chapter 6

Topology on the Real Line

We are discussing in this chapter some basic and important concepts from Topology which are traditionally included in the courses of Advanced Calculus. These include topological and metric spaces in general, and topologies on R in particular, sequences in R, convergent sequences in R, and Cauchy sequences in R. Our discussions are rigorous, but quite introductory to this circle of concepts and ideas. We will give several examples of topological spaces, but our emphasis will be on studying the classical topology on the set of real numbers. Our main tool in doing so will be sequences of real numbers, but whenever possible, and pedagogically feasible, we will use methods applicable in greater generality as well. These latter will give the idea of how the facts and concepts related to the topology of the real numbers can be generalized to other topological spaces. There are two important results in this chapter: Bolzano-Weierstrass’ theorem, that every bounded indexed family of real numbers has an accumulation point, and the theorem characterizing the convergent sequences of real numbers as Cauchy sequences. The former theorem will be used in next chapter when we study limits and continuity of functions. The latter theorem will be used as a motivation to give, in the concluding these Lecture Notes chapter, the construction of the real numbers by using Cauchy sequences of rational numbers which is due to Georg Cantor.

6.1 The Classical Topology on R

Speaking of topologies on a set, there are many such that can be deﬁned on R. One of them, the so called classical topology, is an object of considerations in Advanced Calculus. The classical topology on R is deﬁned by using the total order on R, and the function absolute value on R. The most useful properties of this function are collected in the following proposition.

Proposition 6.1.1 The function | • | : R → R has the following properties. (i) Range(| • |) = R+ ∪ {0}, and only 0 ∈ R is mapped to 0; (ii) The function is multiplicative: |x · y| = |x| · |y|; (iii) The function satisfies the triangle inequality: |x + y| ≤ |x| + |y|; (iv) The function satisfies also the inequality: ||x| − |y|| ≤ |x − y|. Proof. Items (i) and (ii) are straightforward from the definition of | • | in the previous chapter. For item (iii), notice that if x and y have the same sign, that is, they are both non-negative, or both negative, then |x + y| = |x| + |y|. Finally, if x and y have opposite signs, and say |x| ≤ |y|, then |x + y| = |y| − |x| < |y| + |x| (why?) which completes the proof of (iii). The proof of (iv) follows from (iii). Indeed, we have by (iii) that |y| = |x + (y − x)| ≤ |x| + |y − x| and |x| = |y + (x − y)| ≤ |y| + |x − y| which give us that −(|x| − |y|) ≤ |y − x| and |x| − |y| ≤ |x − y|. The last two inequalities are equivalent to ||x| − |y|| ≤ |x − y|. 2

89 Exercise 6.1.1 Give all details in the proof of item (iii) above. Note that the items (i) and (iv) can be used to prove the following.

Proposition 6.1.2 The function d : R × R → R deﬁned by d(x, y) = |x − y| has the properties. (d1) (∀x, y ∈ R)(d(x, y) ≥ 0 ∧ (d(x, y) = 0 ⇔ x = y)). That is, d is positive; (d2) (∀x, y ∈ R)(d(x, y) = d(y, x)). That is d is symmetric; (d3) (∀x, y, z ∈ R)(d(x, y) + d(y, z) ≥ d(x, z)). That is d satisﬁes the triangle inequality. Exercise 6.1.2 Prove the preceding proposition.

The function d makes out of R a very special set - a metric space.

Deﬁnition 6.1.1 A pair (X, d) where X is a set and d : X × X → R is called a metric space, with a metric d if the function d has the properties (d1), (d2), and (d3), that is, if it is positive, symmetric, and satisﬁes the triangle inequality.

It is this metric on R which defines the classical topology on it. The following definition introduces concepts needed to define thisl topology.

Deﬁnition 6.1.2 (1) For a, b ∈ R, a < b, deﬁne the open intervals in R by (a, b) = {x ∈ R | a < x < b }, (a, ∞) = {x ∈ R | a < x }, (−∞, b) = {x ∈ R | x < b }; (2) Let > 0 be a real number, and a ∈ R. Call (a − , a + ) the epsilon nbhd of a; (3) Let U ⊆ R. Call U a nbhd of a ∈ R if (a − , a + ) ⊆ U for some positive ; (4) The subset U ⊆ R is called an open subset of R if it is a nbhd of each one of its points.

Example 6.1.1 Every epsilon nbhd is an open subset of R. Indeed, let b ∈ (a − , a + ). We have that a − < b < a + . Let 1 = min{b − (a − ), (a + ) − b}. Obviously, 0 < 1 ≤ . We claim that

(b − 1, b + 1) ⊆ (a − , a + ).

To this end, let x ∈ (b − 1, b + 1). This means that b − 1 < x < b + 1 or, equivalently that |x − b| < 1. We have now

|x − a| = |(x − b) + (b − a)| ≤ |x − b| + |b − a| < 1 + |b − a|.

By the deﬁnition of 1 we have that

1 ≤ b − a + and 1 ≤ a + − b which amounts to a − b ≤ − 1 and −(a − b) ≤ − 1, and ultimately to |a − b| ≤ − 1. To compete our argument, observe that |x − a| < 1 + |b − a| ≤ 1 + ( − 1) = , and so x ∈ (a − , a + ). 2

Exercise 6.1.3 (1) Verify that - neighbourhoods can be deﬁned in terms of the metric d on R: (a − , a + ) = {x ∈ R | d(x, a) < }. (2) Prove that (i) The empty subset of R, and R itself are open subsets of R; (ii) The union of any family of open subsets of R is an open subset of R; [Let A = {Uλ | λ ∈ Λ} be a family of open subsets of R, and let x ∈ ∪ A = ∪λ∈ΛUλ. We want to show that, for some > 0, we have (x − , x + ) ⊆ ∪ A. But this is obviously the case, because there is a λ0 such that x ∈ Uλ0 , and since Uλ0 is an open subset of R, there is an > 0 such that

(x − , x + ) ⊆ Uλ0 ⊆ ∪A.] (iii) The intersection of any ﬁnite family of open subsets of R is an open subset of R. [Let U1,...,Un be open subsets of R, and let x ∈ U1∩· · ·∩Un. Since, for every i = 1, . . . , n, the n n set Ui is open, there is an i > 0 such that (x−i, x+i) ⊆ Ui. Therefore, ∩i=1(x−i, x+i) ⊆ ∩i=1Ui. n we complete the argument noticing that, for = min{1, . . . , n}, we have (x − , x + ) ⊆ ∩i=1Ui.]

90 We deﬁne next the concept of a topological space. The previous exercise will assure then that the open subsets of R deﬁned above constitute a topology on R.

Deﬁnition 6.1.3 Let X be a set, and let TX = {Uλ | λ ∈ Λ } ⊆ P(X). We say that TX is a topology on X, and that the elements of TX are the open subsets of X if (i) ∅,X ∈ TX ; (ii) If U1,...,Uk ∈ TX , then U1 ∩ · · · ∩ Uk ∈ TX ; (iii) If A ⊆ TX , then ∪A ∈ TX . The pair (X, TX ) is called then a topological space.

When there is no risk of confusion about who the family TX is we abbreviate (X, TX ) to X.

Example 6.1.2 (1) By the definition of topology on X, the sets ∅ and X are open. In fact, the collection TX = {∅,X} trivially satisfies (i), (ii), and (iii) from the Definition above, and is, therefore, a topology on X. Having the minimum possible open sets, this topology is not interesting for applications, and is rightfully called the trivial topology on X . (2) On the other extreme, the collection TX = P(X) also satisfies the conditions (i), (ii), and (iii) from the Definition of topology on X. If the trivial topology is the ”smallest” topology on X, the topology TX = P(X) is the ”largest’ possible one in which every subset of X is open. It is straightforward that this topology is characterized by having every singleton subset {x} ⊆ X, where x ∈ X, is an open subset of X. For this reason, this topology is called the discrete topology on X.

Now we are formally introducing the classical topology on R.

Deﬁnition 6.1.4 The classical topology on R is deﬁned by

TR :=: C := {U ⊆ R | (∀x ∈ U)(U is a nbhd of x) } ⊆ P(R). Vista: As we have seen already, the classical topology on R is determined by the metric on R. Indeed, recall that a set is open in that topology, if every point of that set has an -nbhd contained in the set, and that the -nbhds are of the form U = {x | x ∈ R ∧ d(x0, x) < }. Because of this last description, the classical topology on R is called metric topology. Moreover, it is true that every metric space determines its metric topology, and a topological space (X, TX ) whose topology is determined by a metric d on it is called metrizable. Metrizable topological spaces enjoy very nice properties, and were considered as a (dream-) model of what nice and useful topological space should look like. Some of the properties of the metrizable classical topology of R are studied below. 2

Exercise 6.1.4 Prove that if a 6= b are two real numbers, then there are open subset a ∈ Ua and b ∈ Ub of R, such that Ua ∩ Ub = ∅.

Deﬁnition 6.1.5 A topological space (X, TX ) is called Hausdorﬀ, or separated, (or a T2-space) 0 00 0 00 if for every x1 6= x2 ∈ X there are open subsets x1 ∈ U and x2 ∈ U such that U ∩ U = ∅.

Example 6.1.3 The set of real numbers with the classical topology, (R, C), is a separated topological space. 2 Not every topological space is Hausdorﬀ. Moreover, some of very useful and ”natural” topological spaces are not separated. The example below explains a topology on R which is important for such areas of math as Algebraic Geometry, and which is not Hausdorﬀ.

Example 6.1.4 (Zariski topology on R). By deﬁnition, the open sets in this topology are all subsets U ⊆ R such that R \ U is either ﬁnite, or the whole R. The open sets of this topology are called Zariski open subsets. Let’s verify that this is a topology on R. It is immediate that ∅ and R are open in this topology. Suppose U, V are Zariski open subsets of R. Since R \ U ∩ V = R \ U ∪ R \ V

91 is either finite or the whole R (why?), then, by an easy induction, the intersection of any finite family of Zariski open subsets of R is Zariski open. Finally, for any family U = {Uλ | λ ∈ Λ} of Zariski open subsets of R, we have R \ ∪ U = ∩{R \ Uλ | λ ∈ Λ} which is an intersection of sets such that either all they are equal to R or at least one of them is finite. Therefore, the complement of ∪ U is either R or a finite set, and so, ∪ U is Zariski open. A very important feature of the Zariski topology on R is that every two non-empty Zariski open subsets intersect non-trivially

U 6= ∅,V 6= ∅ (Zariski open) ⇒ U ∩ V 6= ∅.

(Indeed, R \ U ∩ V = R \ U ∪ R \ V 6= R, because it is a ﬁnite set(why?).) Therefore, the Zariski topology is not separated. Or, in other words, ”Zariski is not Hausdorﬀ”. 2

A very important, and closely related to the topology of a topological space, are the closed subspaces.

Deﬁnition 6.1.6 Let (X, TX ) be a topological space. The subset V ⊆ X is called a closed subset of X if X \ V ∈ TX .

Remark 6.1.1 The notion of closed set is not the negation of the concept of open set. So, in general a subset of a topological space can neither be open, nor closed! Moreover, usually most of the subsets are neither open nor closed! For instance the subset (a, b] = { x ∈ R | a < x ≤ b } is neither closed nor open in R (prove that!). 2 It is obvious that knowing the open subset of a topological space X is equivalent to knowing the closed subsets. Therefore the topology of X should be describable in terms of closed subsets only. The corresponding description is given in item (iv) of the exercise below.

Exercise 6.1.5 1) Show that every one-point subset of the R is closed in the classical topology. Conclude that every finite subset of R is a closed subset in the classical topology. 2) A topological space (X, TX ) the one-point subsets of which are closed is called Frećhet (or a T1-) topological space. Prove that every Hausdorff space is a Frećhet space. Conclude by 1) then that (R, C) is a Frećhet space. 2’) Prove that the Zariski topology on R is Frećhet. Conclude that there are Frećhet topological spaces which are not Hausdorff. ”Frećhet is not Hausdorff” either. 3)Prove that (X, TX ) is a Frećhet topological space if, and only if,

(∀x, y ∈ X)((x 6= y) → (∃ U, V ∈ TX )((x ∈ U ∧ y∈ / U) ∧ (x∈ / V ∧ y ∈ V ))

4) Prove, for a topological space (X, TX ), that (i) The empty set, ∅, and X are closed in X; (ii) If D is a finite family of closed subsets of X, then ∪D is a closed subset of X; (iii) If B is a non-empty family of closed subsets of X, then ∩B is a closed subset of X. 5) Let X be a set, and let F ⊆ P(X) have the properties (1), (ii), and (iii) from item 4) of this exercise. Prove that the family A = {X \ V | V ∈ F } defines a topology TX on X such that F consists of the closed subsets of X. Conclude that a topology of a set X can be defined by exhibiting a family of subsets of X satisfying (i), (ii), and (iii) from 4) above. 6) Let X be a set, and let F ⊆ P(X) be defined as follows

F = {Y ⊆ X | Y = X ∨ Y is finite }.

Prove that X has a topology whose closed subsets constitute F. This topology is known as the ﬁnite complement topology on X . (Note that the Zariski topology on R is the ﬁnite complement topology on R.)

92 6.2 Sequences of Real Numbers

This subsection treats rigorously concepts that are, presumably, known to the students from the courses of Calculus... All these concepts are important, because they oﬀer an useful way to describe the classical topology on R. Most of the deﬁnitions we give and the propositions we prove can be generalized, word for word, to any metric space.

6.2.1 Convergent Versus Cauchy Sequences I We begin with a general deﬁnition.

Deﬁnition 6.2.1 A sequence of elements of a set X, or just a sequence in X, is called any function f : N → X. A sequence in X is called stationary or constant if the function f is a constant function

We abbreviate the notation of a sequence: instead of f : N → X, we write < xn > where xn = f(n). Sequences can be considered in an equivalent way as indexed families with indexing set N:

< xn >= {xn | n ∈ N }.

So, two members of the sequence < xn > are distinct if their indices are distinct. In this course, we will be interested in sequences in R.

The most important sequences for us will be the convergent to real numbers ones.

Deﬁnition 6.2.2 Let L be a real number, and < xn > be a sequence of real numbers. We say that < xn > converges to L, and that L is a limit of < xn >, if

(∀ > 0)(∃n0 ∈ N)(∀n ∈ N)(n0 ≤ n → |xn − L| < ).

We write in such a case: < xn >→n L.

Exercise 6.2.1 1) Prove that every stationary sequence in R is convergent. 2) Prove that the sequence f : N → R defined by f(n) = 1/2n converges to 0. Exercise 6.2.2 Prove that, in the definition of convergent sequence, the -nbhd of L can be replaced by any nbhd of L. More precisely, prove that the sequence < xn > is convergent if, and only if, there is a real number L with the following property: for every nbhd U of L, there exists an n0 ∈ N such that xn ∈ U for all n ≥ n0. The last exercise shows how the concept of convergent sequence of real numbers can be generalized to any topological space: the element y ∈ X is a limit of the sequence < xn >, with x ∈ X for all n ∈ N, if every nbhd U of y contains all but finitely many members of the sequence.

It is very important that R is a separated topological space. Because of this, a convergent sequence has a unique limit.

Proposition 6.2.1 Suppose < xn > is a sequence in R. If < xn >→n L1, and < xn >→n L2, then L1 = L2.

Proof. Assume, by RAA, that L1 6= L2, denote by = |L1 − L2|/2. By the deﬁnition of limit, applied to L1 and L2, for this there are natural numbers n1, n2 ∈ N such that

(∀n ∈ N)(n ≥ n1 → |xn − L1| < ) and (∀n ∈ N)(n ≥ n2 → |xn − L2| < ).

Choose any natural number n > max{n1, n2}. We have

2 = |L1 − L2| = |L1 − xn + xn − L2| ≤ |L1 − xn| + |xn − L2| < + = 2, and therefore: 2 < 2 - a contradiction. 2

93 Example 6.2.1 To see how important the Hausdorﬀ condition for the concept of limits of sequences is, consider the following ”pathological” example. Suppose < xn > is a sequence in R consisting of pairwise distinct elements: xn 6= xm for every m 6= n. Then, every real number is a limit of this sequence in the Zariski topology on R! In other words, in the Zariski topology,

∀L ∈ R (< xn >→n L).

Indeed, every open nbhd U of L is non-empty (for L ∈ U), and hence R \ U = {r1, . . . , rk} is a finite set. This means that {xn | n ∈ N} ∩ R \ U is finite, and, therefore, U does not contain no more than finitely many members of < xn >. This means that there is a natural number N such that for all n ≥ N we have xn ∈ U. We know that the Zariski topology on R is not Hausdorff, but is Fréchet. This example shows therefore that the concept of convergent sequences in Fréchet topological spaces is not an interesting concept: it may have examples of ”pathologically looking” sequences. 2 We worked a lot to construct the real numbers as a number system with continuous relation on it. Here is the first benefit we can get from that: a sufficient condition for a sequence to have a limit.

Deﬁnition 6.2.3 (1) The sequence < xn > is called increasing, respectively decreasing, if for every n < m ∈ N we have xn ≤ xm, respectively xm ≤ xn. (2) The sequence < xn > is called monotone if it is either an increasing or a decreasing sequence. A sequence is called strictly increasing/decreasing/monotone, if the inequalities in (1) and (2) are all strict. Here is the important result.

Theorem 6.2.2 If the sequence < xn > is monotone and bounded, then it is convergent, that is, there is a real number L such that < xn >→n L. Proof. If the sequence is, say, increasing, then it is bounded above (because the sequence is bounded), and therefore has a l.u.b. L in R. We are showing next that < xn >→n L. To this end, let > 0. The number L − , being smaller than the least upper bound of the sequence, is not an upper bound of that sequence. This means that there is a natural number n0 such that

L − < xn0 <≤ L. Since the sequence is increasing, for all n0 ≤ n ∈ N we have xn0 ≤ xn and therefore

L − < xn0 ≤ xn ≤ L < L + .

These inequalities for xn can be rewritten as |xn − L| < . The proof is completed. 2

Convergent sequences (of real numbers, in the classical topology) are really special. They have the following feature.

Proposition 6.2.3 If < xn > is convergent (with limit L), then

(∀ > 0)(∃n0 ∈ N)(∀m, n ∈ N)(m, n ≥ n0 → |xn − xm| < ).

Notice the important fact that L does not take part in the feature of < xn >!

Proof. The results is based on the triangle inequality: for every n, m ∈ N

|xn − xm| = |(xn − L) + (L − xm)| ≤ |xn − L| + |xm − L|.

If we can make the summands on the RHS be less than /2, we will be done. But arranging the latter is not a problem, since < xn >→n L. 2

It turns out, sequences with the feature just described are the most important ones in Calculus! They bare the name of the guy who introduced them... long ago.

94 Deﬁnition 6.2.4 The sequence of real numbers < xn > is called a Cauchy sequence if

(∀ > 0)(∃n0 ∈ N)(∀m, n ∈ N)(m, n ≥ n0 → |xn − xm| < ).

So, by the proposition above, every convergent sequence or real numbers is Cauchy! The truth of the matter is that, conversely: Every Cauchy sequence of real numbers has a limit in R. This fact is based on an important property of the classical topology of R, called Bolzano-Weierstrass’ theorem. To formulate this theorem, we need a concept which makes many of the arguments used more transparent: accumulation points.

6.2.2 Accumulation Points of Sets

Definition 6.2.5 Let A ⊆ X where (X, TX ) is a topological space. The point x ∈ X is an accumulation point of A if every open subset U ⊆ X with x ∈ U has infinitely many elements of A, that is the set U ∩ A has infinitely many elements.

Obviously, x is an accumulation point of A if and only if every nbhd of x contains inﬁnitely many elements from A. It is equally obvious that if x is an accumulation point of a subset of A, then it is an accumulation point of A itself. One more thing, which is also obvious, is that ﬁnite sets have no accumulation points!

Example 6.2.2 The two extreme topologies on a set X, namely, the trivial topology and the discrete topology provide examples where the concept of accumulation point is not ”interesting”. What this means is that this concept doesn’t distinguish points in relation with an inﬁnite subset A ⊆ X as in accumulation points of A. And indeed, in the trivial topology, every element of X is an accumulation point, while in the discrete topology no element of X is an accumulation point of A. 2

Example 6.2.3 Our running example of R with the Zariski topology shows that the notion of accumulation points is not interesting even for some (non-trivial) topologies. Indeed, if A ⊆ R is an infinite subset, then every real number L is an accumulation point of A. This is, because every non-empty Zariski open subset of R, in particular, every Zariski open nghd of L, intersects A in infinitely many points! We will show below that, on the contrary, the accumulation points are a fine and useful concept for the classical topology of R. 2 Nevertheless, as we will see shortly, the concept of accumulation point of a set will be very useful in our study of the classical topology on R.

For those who are are uncomfortable with thinking of sets with inﬁnitely many elements, here is a characterization of the accumulation points not appealing to inﬁnite sets. Of course, we have to put a (mild) restricction on the topological space for that.

Exercise-Proposition 6.2.3 Let (X, TX ) be a Fre´chet space, and let x ∈ X. Then x is an accumulation point of A ⊆ X if, and only if, for every open subset U ⊆ X such that x ∈ U, the intersection U ∩ A has at least two elements. In the case of (R, C), this is also equivalent to ”every -nbhd of x contains at least two elements of A”.

Proof One of the directions of the ﬁrst statement is obvious. The other follows by induction, and is based on the fact that, In a Fr´echet space, if U ⊆ X is open, and y ∈ U, then U \{y} is open as well. The second statement, about the space (R, C), is straightforward. 2

The accumulation points of subsets of R have the following characterization in terms of sequences. Sometimes this description helps to prove theorems related to accumulation points in a faster and easier fashion.

95 Theorem 6.2.4 Suppose ∅ 6= X ⊆ R, and r ∈ R. Then, r is an accumulation point of X if, and only if, there is a sequence < xn >⊆ X \{r} such that < xn >→n r. Proof. (⇐) Assuming a sequence of elements of X, every member of which is distinct from r, has a limit r, we have to show that every nbhd of r has at least two elements of X. As usual, it is enough to prove that for nbhds of the type (r − , r + ). To this end, let > 0 be ﬁxed. By the deﬁnition of limit, there is a natural number n0 such that for every natural number n ≥ n0 we have xn ∈ (r − , r + ), or equivalently |r − xn| < . In particular xn0 ∈ (r − , r + ). Denote by

1 = |r − xn0 |. This number is strictly positive, because all members of the sequence are distinct from r. We have also that 1 < . Now, there is a natural number n1 such that for every natural n ≥ n1 we have xn ∈ (r − 1, r + 1). In particular xn1 ∈ (r − 1, r + 1). From this it follows that xn0 6= xn1 and that xn0 , xn1 ∈ (r − , r + ). (⇒) Knowing that r is an accumulation point of X, we want to construct a sequence {xn | n ∈ N} ⊆ X \{r} such that < xn >→n r. We prove here a bit more: the members of the sequence are pairwise distinct real numbers. Consider, for every n ∈ Z+, the nbhds Un = (r − 1/n, r + 1/n) of r. For every n, Un ∩ X has inﬁnitely many elements. Recursively, having chosen xk−1 ∈ Uk ∩ (X \{r}) for k = 0, 1, . . . , n, we choose xn ∈ Un+1 ∩ (X \{r, x0, . . . , xn−1}. By construction, the sequence {xn | n ∈ N} consists of pairwise disjoint elements. Proving that < xn >→n r is left as an exercise. 2

Exercise 6.2.3 Prove that the sequence constructed in part (⇒) of the proof above is convergent with a limit r.

As a ﬁrst application of the previous theorem, we are giving a description of the closed subsets of R.

Theorem 6.2.5 The subset V ⊆ R is closed if, and only if, V contains all its accumulation points (∀x)((x − accumulation point of V ) → (x ∈ V )).

Proof. We have to prove two things: (1) If V is closed, then it contains all its accumulation points, and (2) If V contains all its accumulation points, then V is closed in R. To prove (1), and arguing by contradiction, assume r∈ / V is an accumulation point of V . Since V is closed in R, its complement R \ V is open, and is a nbhd of r. By the previous theorem, there is a sequence < vn >⊆ V with limit r. But then every nbhd of r should contain (even inﬁnitely many) elements of the sequence. The last is not true for R \ V - a contradiction. This proves item (1). To prove (2), arguing again by contradiction, assume V is not closed. So, R \ V is not open, that is, there is a point x ∈ R \ V for which R \ V is not a nbhd. This means in turn that, for every positive integer n, the intersection V ∩ (x − 1/n, x + 1/n) is non-empty. Deﬁne a sequence < vn > by v0 = v1 ∈ V ∩ (x − 1, x + 1) and for n ≥ 2, vn ∈ V ∩ (x − 1/n, x + 1/n). Obviously, < vn >→n x and, since x∈ / V , < vn >⊆ V \{x}. By the previous theorem, x is an accumulation point of V , and therefore, by our hypothesis about V , it belongs to V . So, x∈ / V ∧ x ∈ V - an absurd. 2

The following is an important property of the classical topology on R.

Theorem 6.2.6 (Bolzano-Weierstrass (for sets of real numbers)) Every inﬁnite and bounded subset A ⊆ R has an accumulation point. Proof. Since A is bounded, we have that A ⊆ [a, b] for some real numbers a, b. The idea of the proof is to construct two sequences, one of which, < an >, being increasing, and the other, < bn >, decreasing, and having the following properties.

(∀n, m ∈ N)((an ≤ bm), b − a (∀n ∈ ) b − a = , N n n 2n

96 and (∀n ∈ N)(A ∩ (an, bn) is inﬁnite). Suppose, we have constructed these sequences. Both they are monotone and bounded, so, they are convergent < an >→n L1 and < bm >→m L2.

Since the sequence < an > consists of lower bounds of the sequence < bm >, and since L1 is the least upper bound of {an | n ∈ N}, we immediately get that L1 is a lower bound of {bm | m ∈ N}. Analogously, L2 is an upper bound of {xn | n ∈ N}. From this we get that L1 ≤ L2. On the other hand, since for every n ∈ N b − a 0 ≤ L − L ≤ b − a = , 2 1 n n 2n

we get that L1 = L2(=: L). We claim now that L is an accumulation point for A. Indeed, let > 0 be given. Since L − = L1 − is not an upper bound of {an | n ∈ N} there is n1 ∈ N such that

L − < an1 ≤ L. Analogously, L + = L2 + is not a lower bound of {bm | m ∈ N}, there is n2 ∈ N such that L ≤ bn2 < L + . Denoting by n0 the maximum of n1 and n2, we get that

L − < an0 ≤ L ≤ bn0 < L + .

In other words, (an0 , bn0 ) ⊂ (L − , L + ). But then

A ∩ (an0 , bn0 ) ⊂ A ∩ (L − , L + )

which immediately implies that the -nbhd of L has infinitely elements of A. This completes the proof that L is an accumulation point of A. We are constructing the sequences < an > and < bm > now. We do this recursively. 0 Define a0 = a and b0 = b. We obviously have that 0 < b0 − a0 = (b − a)/2 and A ∩ (a0, b0) = A is an infinite set. Suppose we have defined an and bn such that b − a a ≤ a < b ≤ b , b − a = , and A ∩ (a , b ) is an infinite set. n−1 n n n−1 n n 2n n n

We have [an, bn] = [an, (an + bn)/2] ∪ [(an + bn)/2, bn] and so,

a + b a + b A ∩ [a , b ] = (A ∩ [a , n n ]) ∪ (A ∩ [ n n , b ]) is an inﬁnite set. n n n 2 2 n

an+bn Define an+1 and bn+1 as follows. If A∩[an, 2 ] is infinite, then an+1 = an and bn+1 = (an +bn)/2. an+bn If A ∩ [an, 2 ] is finite, denote an+1 = (an + bn)/2 and bn+1 = bn. It is straightforward to check that b − a a ≤ a < b ≤ b , b − a = , and A ∩ (a , b ) is an infinite set. n n+1 n+1 n n+1 n+1 2n+1 n+1 n+1

Using induction, we see that the constructed sequences < an > and < bm > have the promised properties. The proof of the theorem is complete. 2

Exercise 6.2.4 (∗) Here is another, more direct, way of proving the Bolzano-Weierstrass Theorem. (1) Deﬁne Σ = {r | (−∞, r] ∩ A is finite} ⊆ R. Prove that ∅= 6 Σ, and that Σ is bounded above. (2) Denote by s the least upper bound of Σ, and prove that, for every > 0, the set (s, s + ) ∩ A is inﬁnite. Conclude that s is an accumulation point of A.

Bolzano-Weierstrass theorem is instrumental in proving many results in Calculus. To see it work, we need to extend the concept of accumulation point from sets to multisets of real numbers, and in particular - to sequences.

97 6.2.4 Accumulation Points of Sequences We use here ”multiset” and ”indexed family” interchangeably. As we know, they mean the same thing.

Deﬁnition 6.2.6 Let A = {xλ ∈ R | λ ∈ Λ} be an indexed family of real numbers, and let L ∈ R. The number L is called an accumulation point of A if for every nbhd U of L we have that

{λ ∈ Λ | xλ ∈ U} is an infinite set. A bit informally, L is an accumulation point of an multiset if every nbhd of L contains infinitely many members of the multiset. Obviously, for A to have an accumulation point, Λ has to be infinite.

The accumulation points of multisets can be studied in utmost generality, for any inﬁnite indexing set Λ, but in our course we are interested in the smallest possible inﬁnite such set: Λ ⊆ N. So, from this point on, we consider sequences or real numbers only.

First we restate the deﬁnition of an accumulation point of a sequence.

Proposition 6.2.7 The real number L is an accumulation point of the sequence < xn > if, and only if,

(∀ > 0)(∀n ∈ N)(∃n0 ∈ N)(n0 > n ∧ |xn0 − L| < ).

Proof (⇒) Assume L is an accumulation point of < xn >. Every nbhd of L contains inﬁnitely many members of < xn >. Note that ”inﬁnitely many members of < xn > are in the nbhd” is

equivalent to ”for every n ∈ N there is n0 > n such that xn0 is in that nbhd. So, in particular, for every > 0 the nbhd (L − , L + ) of L has this property. (⇐) The claim follows from the fact that every nbhd of L, by deﬁnition of nbhd, has a subset of the form (l − , L + ) for an appropriate > 0. 2

It is obvious that the limit of a convergent sequence is an accumulation point of that sequence. The opposite is not true in general: a real number may be an accumulation point without being a limit of a sequence as the following example shows.

Example 6.2.4 Consider the sequence < xn > be deﬁned by the function f : N → R where f(2n) = n and f(2n + 1) = 1/(n + 1). The real number L = 0 is an accumulation point of < xn >, but is not its limit. 2

Exercise 6.2.5 Prove the claim about L = 0 in the example above. Prove also that < xn > has only one accumulation point, and that it does not have a limit. We will characterize shortly the convergent sequences in terms of accumulation points only. Here is a ﬁrst useful claim.

The fact that N is totally ordered helps us deﬁne sub-families of sequences which are sequences as well. Recall that sequences are just functions with domain N and co-domain R.

Deﬁnition 6.2.7 The sequence g : N → R is a subsequence of f : N → R if there is a strictly increasing function h : N → N such that g = f ◦ h.

In other words, g(k) = f(h(k)) for all k ∈ N. Denoting nk = h(k), and using the standard notations

xn = f(n), and g(k) = yk, we get yk = xnk . So that a subsequence of < xn > is a sequence < xnk > where n0 < n1 < ··· < nk < ··· are natural numbers.

The following proposition is the counterpart, in the case of sequences in R, of the theorem characterizing the accumulation points of sets in R via sequences in the sets converging to those points.

98 Proposition 6.2.8 The real number L is an accumulation point of the sequence < xn > if, and

only if there is a subsequence < xnk > of < xn > which converges to L.

Proof Let < xn > be defined by f : N → R. To prove the ”only if” part of the claim, assume L is an accumulation point of < xn >. Consider the set A = {n ∈ N | xn = L }. We consider two cases: A is finite, and A is infinite. For the latter case notice that A ⊆ N, as well as A \ B ⊆ N have least elements for every finite subset B ⊂ A. Define the function g : N → N recursively as follows. Define g(0) to be the least element of A, and, having defined g(0), g(1), . . . , g(k), set g(k + 1) to be the least element of A \{g(0), . . . , g(k)}. Obviously, the function g is strictly monotone, and therefore

defines a subsequence < xnk > of < xn > by nk = f(g(k)). The sequence < yk > with yk = xnk , being stationary (i.e., yk = L for every k), converges to L. For the case when A is finite, L - an accumulation point of < xn > is equivalent to L being n an accumulation point of Ran(f). Let n = 1/2 for n ∈ N. Consider the sets Bn = {m ∈ N | xm ∈ (L − n,L + n) } where n ∈ N. All these sets are infinite, and Bn+1 ⊆ Bn for every n ∈ N (Why?). Define the function h : N → N recursively as follows. Set h(0) to be the least element of B0, and, having defined h(0), h(1), . . . , h(k), define h(k + 1) to be the least element of Bk+1 \{h(0), h(1), . . . , h(k)}. It is straightforward by the construction used that h is a strictly monotone function. Then the sequence < zk > corresponding to f ◦ h : N → R is a subsequence of < xn >, and satisfies the relation 1 |z − L| < = k k 2k and therefore converges to L. This completes the proof of the ”only if” part of the proposition. For

the ”if” part, notice that if L is a limit of a sub-sequence < xnk > of < xn >, then every nbhd of L contains all but ﬁnitely many members of the subsequence, and therefore contains inﬁnitely members of the sequence < xn >. So, L is an accumulation point of < xn >. 2

Following is the promised Bolzano-Weierstrass theorem for accumulation points of multisets. We will apply this theorem in the realm of sequences, but will prove it in the general set-up 0f indexed families of real numbers - the proof in the general case is not more complicated than the one for sequences!

Theorem 6.2.9 (Bolzano-Weierstrass (for multisets of real numbers)) If the multiset of real numbers A is inﬁnite and bounded, then it has an accumulation point. In other words, if A = {xλ | λ ∈ Λ} is an indexed family such that Λ is an inﬁnite set, and there are m, M ∈ R so that for every λ ∈ Λ we have m ≤ xλ ≤ M, then A has an accumulation point.

Proof Let the multiset A be defined by the function f :Λ → R. There are two cases to consider: Ran(f) finite, and Ran(f) infinite. In the former case, Ran(f) = {L1,...,Ls}. Denote by Λi = {λ ∈ Λ | xλ = Li }. Since obviously Λ = Λ1 t · · · t Λs, at least one of the sets Λi, i = 1, . . . , s is infinite, and so one of the numbers L1,...,Ls is an accumulation point of A: every nbhd of this Li has infinitely many members of the family. In the latter case for Ran(f), since it is bounded, with the help of Bolzano-Weierstrass theorem we get a real number L which is an accumulation point of Ran(f), and therefore - an accumulation point of A. 2

6.2.5 Convergent Versus Cauchy Sequences II We are completing in this subsection the comparison between convergent and Cauchy sequences of real numbers. Before that we give, as promised earlier, a characterization of convergent sequences in terms of accumulation points only.

Theorem 6.2.10 A sequence is convergent if, and only if, it is bounded and has a unique accumulation point.

Proof (⇒) Let < xn >→ L. We show that L is the only accumulation point of < xn >. Indeed, by RAA, if L0 is another one, let = |L − L0|/2 > 0. By the deﬁnition of limit, there are only ﬁnitely

99 many members of < xn > outside (L − , L + ). Therefore, there are no more than finitely many 0 0 0 members of < xn > in (L − , L + ) which contradicts the fact that L is an accumulation point of < xn >. On the other hand, as we already know, a convergent sequence is bounded. The proof of the only if part of the theorem is complete. (⇐) Conversely, let < xn > be a bounded sequence which has a unique accumulation point L. We will show that < xn >→ L. To this end, observe first that, for every > 0, the set A = {xn | xn ∈/ (L − , L + ) } is bounded. If this set is infinite, then it would contain a subsequence of < xn > which, being bounded as a part of < xn >, would have an accumulation point 0 0 L itself. The number L is clearly an accumulation point of < xn > as well. Therefore, we should 0 0 have L = L. But L ∈/ (L − , L + ), and can not be equal to L - a contradiction! So, A is finite for every > 0. But this is equivalent to having < xn >→n L. 2

We are closing this subsection by proving that convergent sequences of real numbers and Cauchy sequences is the same thing.

Theorem 6.2.11 A sequence < xn > in R is convergent if, and only if, it is Cauchy sequence.

Proof. The ”only if” part of the theorem was proved earlier. For the ”if” part, assume < xn > is a Cauchy sequence. Since every Cauchy sequence is bounded, it has an accumulation point. To complete the proof, we have to only show that a Cauchy sequence has only one accumulation 0 point. Arguing by RAA, assume L and L are two distinct accumulation points of < xn >. By the deﬁnition of Cauchy sequence we have that

(∀ > 0)(∃n0)(∀m, n)((m, n ≥ n1) → (|xm − xb| < )). Also, by the characterization of the accumulation points of sequences we have

0 (∀ > 0)(∀m0)(∃n, m)(n, m > m0 ∧ |L − xn| < ∧ |L − xm| < ).

0 To get a contradiction, let = |L − L |/3, and let n0 be chosen as above. Choose m0 = n0 and let 0 m, n > n0 such that ∧ |L − xn| < ∧ |L − xm| < . We have, using the triangle inequality, that

0 0 3 = |L − L | = |L − xn + xn − xm + xm − L |

0 ≤ |L − xn| + |xn − xm| + |xm − L | < + + . So, 3 < 3 - an absurd. This completes the proof of the ”if” part, and of the theorem. 2

6.3 Arithmetic Operations and a Relation on Sequences

The convergent sequences have many of the properties of the real numbers: we can add, multiply and compare sequences pretty much the way we do with real numbers. As a direct consequence, we can introduce subtraction and division of (convergent) sequences as well. We will see that the arithmetic operations and the relation between convergent sequences can be translated into the usual arithmetics operations and relation (≤) between their limits. It is this translation which lies in the basis of Cantor’s construction of the reals...

We know that sequences of real numbers are just functions from N to R. Since R is a ring, we can deﬁne operations between sequences of real numbers. As a matter of fact, the domain, N, of these functions doesn’t matter in this deﬁnition! We can do the same for functions with values in R (real valued functions) and with any, non-empty, domain X. The set of these functions is denoted by F(X, R).

Deﬁnition 6.3.1 Deﬁne operations addition, +, and multiplication, ·, on F(X, R) as follows. Let f, g ∈ F(X, R). (1) f + g : X → R is such that (∀x ∈ X)((f + g)(x) = f(x) + g(x)) (2) f · g : X → R is such that (∀x ∈ X)((f · g)(x) = f(x) · g(x)).

100 Exercise 6.3.1 Prove that (F(X, R), +, ·) is a ring. That is, the operations are commutative and associative, have neutral elements, every element has opposite, the multiplication is distributive over the addition. Show also that unless X is a singleton, (F(X, R), +, ·) is not a ﬁeld.

Deﬁnition 6.3.2 We say that the sequence < xn > is less than or equal to the sequence < yn >, and we write < xn >≤< yn >. if

(∃n0 ∈ N)(∀n ∈ N)(n ≥ n0 → xn ≤ yn).

When X = N, we agreed to denote the functions f : N → R by < xn > where xn = f(n). The propositions bellow are stated in these notations. Thus, if in addition yn = g(n), zn = (f + g)(n), and if wn = (f · g)(n), then < zn >=< xn + yn >, and < wn >=< xn · yn >.

The following theorem establishes the fact of the translation promised above.

Theorem 6.3.1 Suppose < xn >→n L and < yn >→n M. Then we have (1) < xn + yn >→n L + M; (2) < xn · yn >→n L · M; (3) If M 6= 0, then (∃n0)(∀n ∈ N)((n ≥ n0) → (|yn| ≥ |M|/2)); (4) If M 6= 0, and yn 6= 0 for all n ≥ n0, the any sequence < zn > with zn = 1/yn for n ≥ n0 is convergent, and < zn >→n 1/M; (5) If < xn >≤< yn >, then L ≤ M (6) We have

L < M ⇔ (∃ > 0)(∃n0 ∈ N)(∀n ∈ N)((n ≥ n0) → ( < yn − xn));

(7) If L = M, and if the sequence < zn > has the property that

(∃n0 ∈ N)(∀n ∈ N)((n ≥ n0) → (xn ≤ zn ≤ yn)),

then < zn >→n L as well. Proof All items in the theorem, with the exception of (6), are standard, and the proofs are routine. They are left as an exercise. Item (6) is needed in order to introduce relation between Cauchy sequences later on, when we do Cantor’s construction of the reals. We are proving (6) now. Direction (⇒). Let = δ/3. There is an n1 such that |xn − L| < for all n ≥ n1. Similarly, there is an n2 such that |yn − M| < for all n ≥ n2. Denoting n0 = max{n1, n2}, and taking any n ≥ n0, we have

xn < L + = L + (M − L)/3 = (2L + M)/3

and yn > M − = M − (M − L)/3 = (L + 2M)/3. Therefore, for the chosen n, we get

yn − xn > (L + 2M)/3 − (2L + M)/3 = (M − L)/3 = .

In the opposite direction, (⇐), we have, by item (5), first that L ≤ M Assuming, by RAA, that M = L = A, then the sequence < zn > defined by z2n = xn and z2n+1 = yn is convergent with limit A. But then the /2−nbhd of A contains all but finitely many members of < zn >. In particular, for n ≥ n0, we have

|zn − zn+1| = |zn − A + A − zn+1| ≤ |zn − A| + |zn+1 − A| < /2 + /2 = .

This implies that |xn − yn| < for all n ≥ n0 which is not possible due to the inequality < yn − xn for all big n too. This contradiction ﬁnishes the proof of item (6). 2

Exercise 6.3.2 Prove items (1)-(5) and item (7) in the theorem above.

101 Exercise 6.3.3 (1) Show that if < xn >→n L, and if a ∈ R, then < a · xn >→n a · L. (2) If, in addition to (1), < yn >→n M, then < xn − yn >→n L − M. (3) If, in addition to (2), M 6= 0, and n0 ∈ N is such that (∀n ∈ N)(n ≥ n0 → yn 6= 0), then for any sequence < zn > with zn = xn/yn for n ≥ n0, we have < zn >→n L/M.

Remark 6.3.1 As we know, the set F(N, R) is a commutative ring. The theorem above implies that the sub-set of convergent in R, that is, Cauchy, sequences has a structure of a commutative ring, with the inherited from F(N, R) operations, as well. Denote this set by Cauchy(R). Because of just explained property of Cauchy(R), we say that it is a sub-ring of F(N, R). Both rings are not fields: not all non-zero elements of these have reciprocals. Also, the order ≤ is only a partial order. Although, by the cited theorem, the arithmetic properties of Cauchy sequences can be nicely matched with the operations on their limits, the ring Cauchy(R) is far from being a copy of R. But if one identifies the Cauchy sequences having the same limit, then, that’s a theorem now, the resulting set can be identified to R. As we’ll see shortly, Cantor’s construction of R follows this path, but works with Cauchy sequences in Q only. 2 We close this subsection by proving some facts relating properties of sub-sequences of a sequence to the convergence of that sequence.

Theorem 6.3.2 (1) If the sequence < xn > is convergent to L, then every its subsequence < xnk > is convergent to L;

(2) The sequence < xn > is convergent if, and only if, every its subsequence < xnk > is convergent; (3) Suppose that < xn > is bounded, and that all its convergent subsequences have the same limit L. Then < xn >→n L, and therefore, every sub-sequence of < xn > is convergent (to L). (In the hypothesis for (3), we do not impose the restriction that all subsequences be convergent. The restriction is: those subsequences which are convergent all have the same limit.) Proof. The proofs of (1) and (2) are easy, and are left as an exercise. We are proving item (3) now. First oﬀ, since < xn > is bounded, it does have a convergent subsequence. So, the set of convergent subsequences of < xn > is not empty, and, by the hypothesis, all they have the same limit L. If, arguing by RAA, we assume that < xn > is not convergent itself, then there is an −nbhd of L which does not contain inﬁnitely many elements of < xn >. Therefore, there is a subsequence of < xn > outside of that −nbhd of L. The latter subsequence is bounded, as a subsequence of < xn >, and has a convergent subsequence itself. The limit of this subsequence can not be L, because all its elements are no closer than to L. On the other hand, new subsequence is a subsequence of < xn >, and its limit should therefore be L - a contradiction. 2

Exercise 6.3.4 Prove items (1) and (2) of the previous theorem.

6.4 Intro to Cantor’s Real Numbers

One of the morals we can draw from the previous section is that with the operations addition and multiplication, and with the relation ≤ for sequences, the convergent sequences of real numbers share many, if not all, of the properties of real numbers. The link between the two, sequences on one hand and real numbers on other, is given by the limits of the convergent sequences. Having in mind that convergent and Cauchy are synonyms as far as sequences are concerned, we see that Cauchy sequences, with the two operations and the relation above, behave much like real numbers. This observation leads naturally to another way of constructing the real numbers: instead of Dedekind cuts, one, following Cantor, can work with Cauchy sequences of rational numbers. Of course, in this approach one is bound to work with rational numbers only: they are constructing the real numbers anew, and should know nothing about them in this construction.

So, one considers Cauchy sequences of rational numbers, but for that they have to deﬁne these sequences in terms of rational numbers only. This can easily be done by modifying the deﬁnition we introduced in section 6.2:

102 Deﬁnition 6.4.1 The sequence of rational numbers < xn > is called a Cauchy sequence if

(∀ 0 < ∈ Q)(∃n0 ∈ N)(∀m, n ∈ N)(m, n ≥ n0 → |xn − xm| < ).

Now, some of these Cauchy sequences will have a limit in Q, but most of them wont. Also, in the cases when the limit exists, there are many Cauchy sequences with the same limit, and in order to speak of the limits and of sequences interchangeably, one has to identify all Cauchy sequences having the same limit, and associate this class of identiﬁed sequences with their common limit. It turns out that two sequences, < xn > and < yn > have to be identiﬁed if and only if

< xn − yn >→n 0.

Notice that the sequence < xn − yn > is a Cauchy sequence of rational numbers. Since 0 ∈ Q, one can express the condition for identiﬁcation of Cauchy sequences entirely in the realm of Q. A sequence with limit 0 is called a null sequence. So, more formally, one introduces a relation between Cauchy sequences as follows

< xn >∼< yn > if < xn − yn > is null.

This relation turns out to be an equivalence relation on the set Cauchy(Q) of Cauchy sequences of rational numbers. The corresponding quotient set is the set of Cantor real numbers

RC = Cauchy(Q)/ ∼ .

The operations addition and multiplication and the relation ≤ on Cauchy(Q) descend to RC to make out of it an ordered ﬁeld. Also, associating with every rational number x the class of the stationary sequence < xn >, xn = x, one gets a natural map

ψQ : Q → RC which is easily seen to be a monotone homomorphism, with Ran(ψQ) ⊆ RC a dense subset. Finally, and this is the hardest in this approach to prove, one shows that the relation ≤ has the least upper bound property, so ≤ is continuous. This all means that RC is another ordered completion of Q, and therefore, as we already know, is canonically identiﬁable with R.

The full story of the set RC is revealed, naturally, in (the last chapter of) the full version of the LN.

103 Chapter 7

The Sets F(X,Y ),X,Y ⊆ R

The functions we are discussing in this chapter have domains and co-domains subsets of the set of real numbers: f : X → YX,Y ⊆ R. Such functions are called real valued functions of one real variable. Without much loss of generality, we will consider the co-domains to be just R. When the distinction is important, we will consider smaller co-domains as well. Keep in mind though that two functions are equal when they have the same domains, co-domains, and the same values. So, the co-domains do matter! Our ultimate goal here would be to understand as much as possible the properties of continuous functions. But there is an important intermediate step in doing this: understanding functions which have limit at a point. Although the notion of limit of a function is considered as an intermediate step on the way toward continuous functions, it is actually more subtle than continuity. We are discussing the latter using the language of sequences of real numbers. So, the knowledge from Chapter 6 is of great importance here.

7.1 Limits of Functions

We begin with a definition. Notice that limit is defined only for accumulation points of the domain of a function! We begin with an − δ (read as ”epsilon-delta”) definition. There are no sequences involved in it. But we will quickly switch to using those, restating the notion of limits in that language.

Deﬁnition 7.1.1 (Limit of a function. − δ) Let f : X → R with X ⊆ R, and L ∈ R. Let 0 further x ∈ R be an accumulation point of X. We say that f has a limit L at x0, and write limx→x0 f(x) = L, if

(∀ > 0)(∃δ > 0)(∀x ∈ X)(0 < |x − x0| < δ → |f(x) − L| < ).

Important: the number x0 in the Deﬁntion above need not belong to X, but it should be an accumulation point of X. This has to be the case even when x0 ∈ X. The question whether there is a limit of a function at a point which is not an accumulation point of the domain of that function makes no sense!

An instance of x0 being an accumulation point of X is when, for some > 0, the punctured epsilon nbhd of x0, or half of it, is a subset of X:(x0 − , x0 + ) \{x0} ⊆ X, or (x0 − , x0) ⊆ X, or (x0, x0 +) ⊆ X. This are the cases considered in any standard Calculus book. Needless to say, these represent very incompletely the whole variety of possible cases of x0 being an accumulation point of X.

104 Example 7.1.1 Suppose the function f has a domain which is discrete subset of R, that is, suppose Dom(f) has no accumulation points. Then f has no limit at any real number. (We will see shortly that such a function, not having limits at all, is actually continuous).

0 Exercise 7.1.1 (1) Prove that limx→x0 f(x) = L if, and only if, x is an accumulation point of X, and (∀ > 0)(∃δ > 0) such that

0 0 0 P(f)∗ ((x − δ, x + δ) ∩ (X \{x })) ⊆ (L − , L + ).

0 (2) Prove that limx→x0 f(x) = L if, and only if, x is an accumulation point of X, and (∀ > 0) the total pre-image P∗(f)((L − , L + )) is a non-empty (may be punctured) nbhd of x0, that is

P∗(f)((L − , L + )) = X ∩ U

where U is a nbhd of x0 in R. Led by the need for satisfying the restriction 0 < |x − x0| in the deﬁnition of limit, we introduce appropriate sequences to be used in our reasonings below. Such sequences exist only for x0 which are accumulation points of the domain of f.

0 Deﬁnition 7.1.2 Let f : X → R, and x ∈ R. The sequence < xn > is called a good sequence for f at x0 if 0 (1) < xn >⊆ X \{x }; 0 (2) < xn >→n x . Observe that good sequences for f at x0 exist if, and only if, x0 is an accumulation point for X.

0 Exercise 7.1.2 Show that if < xn > and < yn > are good sequence for f at x , then so is the sequence < zn > deﬁned by xk if n = 2k zn = yk if n = 2k + 1 Here is the reformulation of limit at x0 in terms of sequences.

Theorem 7.1.1 (Limit of Function; Sequential Approach) Let f : X → R be a function with x0 ∈ R - an accumulation point of X. 0 (1) Assume that, for every good sequence < xn > for f at x , the sequence < f(xn) > is convergent. Then the limit of any such < f(xn) > is independent of the sequence. 0 (2) limx→x0 f(x) exists if, and only if, for every good sequence < xn > for f at x , the sequence < f(xn) > is convergent. In such a case, we have

< f(xn) >→n lim f(x). x→x0

Proof. For item (1) we have to show that for any good sequence < xn >, the limit of < f(xn) > is equal to L which is the same for all such sequences. To this end, let < f(xn) >→n L1 and < f(yn) >→n L2 for the good sequences < xn > and < yn >. As we know from a previous exercise, the sequence < zn > where z2n = xn and z2n+1 = yn, is a good sequence as well. So < f(zn) >→n L3. We know that a subsequence of any convergent sequence is convergent to the same limit. Since both < f(z2n) > and < f(z2n+1) > are subsequences of < f(zn) >, they are convergent both to L3. This immediately gives that L1 = L2 = L3. Proof of item (2). Assume limx→x0 f(x) = L. We have to show that, for every good sequence of f 0 at x , we have < f(xn) > is convergent. We will prove that actually < f(xn) >→n L. Let > 0. By the deﬁnition of limit of a function at a point we have that there is a δ such that

y ∈ (x0 − δ, x0 + δ) ∩ (X \{x0}) → f(y) ∈ (L − , L + ).

105 0 But since < xn > is a good sequence for f at x , there is n0 such that for n ≥ n0 we have 0 0 xn ∈ (x −δ, x +δ). This immediately implies that, for n ≥ n0, we have f(xn) ∈ (L−, L+). Since this is true for any , we get that < f(xn) >→n L as promised. To prove the other direction in (2), 0 assume that, for every good sequence of f at x , we have the sequence < f(xn) > is convergent. By item (1) we have that all these sequences converge to the same real number, call it L. We will prove now that limx→x0 f(x) = L. Arguing by RAA, assume there is an 0 > 0 such that for every δ > 0 0 0 0 there is an y ∈ (x −δ, x +δ)∩(X \{x }) such that |f(y)−L| ≥ 0. Choosing δ = 1, 1/2,..., 1/n, . . . 0 0 0 we get a good sequence < yn > with yn ∈ (x −1/n, x +1/n)∩(X \{x }). But then, by our assumption, < f(yn) >→n L. On the other hand, for all n, we have |f(yn) − L| ≥ 0. This contradiction ﬁnishes the completes our argument. 2

Knowing the relation between convergent and Cauchy sequences, we can reformulate item (2) of the last theorem as follows.

Corollary 7.1.2 Let f : X → R be a function, and let x0 ∈ R be an accumulation point of X. 0 Then, limx→x0 f(x) exists if, and only if, for every good sequence < xn > for f at x , the sequence < f(xn) > is a Cauchy sequence. Proof Obvious. 2

The two ways of expressing the concept of limit of a function are useful in diﬀerent situations. In what follows, we use the sequential approach to prove properties of functions having limits at a point (the same point for all functions involved!).

Theorem 7.1.3 Let f, g : X → R be two functions, and let x0 ∈ R be an accumulation point of X. Suppose limx→x0 f(x) = L and limx→x0 g(x) = M. We have (1) limx→x0 (f + g)(x) = L + M, (2) limx→x0 (f · g)(x) = L · M, (3) If M 6= 0, then there is a δ > 0 such that (∀x ∈ X)((0 < |x − x0| < δ) → (|g(x)| > |M|/2)), (4) If M 6= 0, then limx→x0 (f ÷ g)(x) = L ÷ M, (5) If there is a δ > 0 such that for all x ∈ X for which 0 < |x − x0| we have f(x) ≤ g(x), then L ≤ M, (6) (Squeeze Theorem) Suppose h : X → R as well, and that there is a δ > 0 such that for all x ∈ X and 0 < |x − x0| < δ we have f(x) ≤ h(x) ≤ g(x). If L = M, then h has a limit at x0, and limx→x0 h(x) = L(= M). Proof. The model of proving these claims is as follows. Choose a good sequence for both f and g at x0, and, using Theorem 7.1.1, reduce proving the corresponding claim to a claim about converging sequences. Items (1), (2), (4), (5), and (6) follow directly from the similar claims about sequences discussed in the previous section. We are doing item (3) directly, without using sequences. Let = |M|/2. By the deﬁnition of limx→x0 g(x) = M we have that there is a δ > 0 such that

0 0 x ∈ (x − δ, x + δ) ∩ (X \{x0}) → g(x) ∈ (M − , M + ).

Notice now that if M > 0, then (M−, M+) = (|M|/2, 3|M|/2), and if M < 0, then (M−, M+) = (−3|M|/2, −|M|/2). Therefore, for all x ∈ (x0 − δ, x0 + δ) ∩ (X \{x0}) we have g(x) > |M|/2 when M > 0, and g(x) < −|M|/2 when M < 0. In both cases |g(x)| > |M|/2. This completes the proof of (3). 2

Exercise 7.1.3 Prove items (1), (2), (4), (5), and (6) of the previous theorem. Functions and sequences which converge to 0 have a very useful for technical purposes property. Here it comes.

Theorem 7.1.4 (1) Suppose < xn > is bounded, and that < yn >→n 0. Then we have < xn · yn >→n 0 as well.

106 (2) Suppose f, g : X → R are such that limx→x0 f(x) = 0, and that g is bounded in a punctured nbhd 0 of x . Then we have limx→x0 (f · g)(x) = 0 as well.

Proof. Item (2) follows immediately from item (1). Item (1) is standard. Both are left as an exercise. 2

Exercise 7.1.4 Prove the previous theorem.

7.2 Compact Subsets of R

The concept of sub-sequences allows us to deﬁne a very special type of widely used subsets of R - the compact subsets. We will have application of these subsets in the section of continuous function when we prove a theorem of Weierstrass.

We know from the theorem of the previous subsection that a sub-set of R is closed if, and only if, it contains all its accumulation points. In particular, if a subset has no accumulation points, then it is closed. This happens when the subset is finite (for only infinite subsets can have accumulation points). Of course, we knew finite sets are closed without appealing to that theorem: this follows at once from the fact every point in R is closed, and that, for every topological space, a union of finite closed subsets is closed. On the other hand, there are infinitely closed subsets of R which do not have accumulation points (for instance, N ⊂ R). By the contra-positive of Bolzano-Weierstrass’ theorem, in the latter case the set should be unbounded. Turns out that the sub-sets of R which are closed and bounded are very important - they are compact!

Deﬁnition 7.2.1 The subset C ⊆ R is called a compact subset if every sequence < xn >⊆ C has

a convergent sub-sequence with limit in C: < xnk >→k c0 ∈ C.

The proof of the following characterization of compact subsets of R uses the Bolzano-Weierstrass theorem in a signiﬁcant way.

Theorem 7.2.1 A set X ⊆ R is compact if, and only if it is closed and bounded.

Proof. We prove ﬁrst that X - closed and bounded in R implies that X is compact. To this end, let f =< xn >⊆ X be a sequence in X. Since X is bounded, the sequence is also bounded, and therefore has an accumulation point x. This x is an accumulation point of X as well, and since X is closed, x ∈ X. By the characterization of the accumulation points of sequences, we get that < xn > has a subsequence converging to L. So, X is compact. The opposite direction, that X compact implies that X is closed and bounded, can be proved as follows. Let y be an accumulation point of X. Then, there is a sequence in Xis sequence has a subsequence which converges to a point x ∈ X. But x being a limit of the subsequence, is an accumulation point thereof, and is therefore an accumulation point of the sequence as well. Since every convergent sequence has a unique accumulation point, we get that x = y, and that y ∈ X. This means that X contains all of its accumulation points, and therefore is closed. To prove that X is also bounded, notice (and prove it!) that an unbounded sub-set of R has a an inﬁnite sub-set without accumulation points, and so no sequence in that sub-set can be convergent. 2

Exercise 7.2.1 Fill in the gaps in the proof of the theorem above.

7.3 Compact Subsets of a Topological Space

One can go even further and show that the compactness can actually be explained without mentioning sequences at all! That is this concepts can be generalized to any topological space. We need a deﬁnition ﬁrst.

107 Definition 7.3.1 Let Y be a sub-set of a topological space (X, TX ). An open cover of Y is a family U ⊆ TX such that Y ⊆ ∪ U. The family V ⊆ TX is called a sub-cover of U if V is an open cover of Y and V ⊆ U. The open cover U of Y is called finite if U is a finite set. We have the following result.

Theorem 7.3.1 A sub-set C of R is compact if, and only if, every open cover of C has a ﬁnite sub-cover.

Proof (⇐) Assume every open cover of C has a finite sub-cover, and {xn | n ∈ N} is a sequence in C. We have to prove that the sequence has an accumulation point in C (this being equivalent to the sequence having a convergent in C sub-sequence). By RAA, assume there is no such a point. This means that every point y ∈ C has a nbhd Uy which contains finitely many members of the sequence. The family {Uy | y ∈ C} is an open cover of C. So, according to our assumption, there are finitely many points y1, . . . , ym ∈ C such that C = Uy1 ∪ · · · ∪ Uym . Therefore every member of the sequence belongs to one of these finitely many sets. Since each set Uyj contains finitely many members of the sequence, it follows that the sequence has finitely many members - an absurd! (⇒) Assuming that C is compact, that is, every sequence in C has a sub-sequence converging in C, and that U = {Uλ | λ ∈ Λ} is an open cover of C, we have to prove that U has a finite subset which is also a cover of C. If U is a finite set, we are done. So, we may assume that U is infinite. We consider first the case U is denumerable: U = {Un | n ∈ N }, and argue by contradiction. So suppose there is no a finite subset of U which is also a cover. This means that for every n1, . . . , nk

Un1 ∪ · · · ∪ Unk $ C.

We are constructing a sequence in C which is not convergent in C. Choose c0 to be any element of C, and let Un0 contain c0. Note that the latter exists, because ∪n∈NUn = C. Choose for

c1 any element of C which is not in Un0 , and let Un1 be such that c1 ∈ Un1 . Having chosen

c0, c1, . . . , cm with the corresponding Un0 ,Un1 ,...,Unm , choose for cm+1 an element of C which is

not in Un0 ∪ Un1 ∪ · · · ∪ Unm , and ﬁnd Unm+1 that contains cm+1. Since for every m ∈ N we have

unm ∩ {cn | n ∈ N } ⊆ {c0, c1, . . . , cn}

we immediately conclude that < cn > is not convergent in C. (ﬁll in the details as an exercise). The most general case, when U is not countable follows from the ”countable” case through the following lemma.

Lemma 7.3.2 If C ⊆ R is a compact sub-set, then every cover of open subsets of C has a countable sub-cover. The proof of lemma is a little bit involved. We will prove it in 4 steps. Step (1): For every positive integer n, there are ﬁnitely many points {c(n), . . . , c(n)} ⊆ C such that 1 kn 1 1 C ⊆ ∪kn c(n) − , c(n) + . i=1 i n i n

(n) Therefore, for every c ∈ C there is a 1 ≤ j ≤ kn so that |c − cj | < 1/n. Observe now that the family of intervals

(n) (n) {Vn,j = (cj − 1/n, cj + 1/n) | 1 ≤ n ∈ N, 1 ≤ j ≤ kn} contains countably many elements. Step (2): Let V ⊆ R be an open sub-set, and let c ∈ C ∩ V . There exist n, j such that

c ∈ Vn,j ⊆ V.

Since c ∈ V , and V is open in R, there is an > 0 such that (c − , c + ) ⊆ V . Since R has the Archimedean property, there is an n ∈ N such that n > 2/. By Step (1), there is a

108 (n) (n) c ∈ (cj − 1/n, cj + 1/n) = Vn,j. We have then that c ∈ Vn,j ⊆ (c − , c + ) ⊆ V (verify this as an exercise!). The proof of Step (2) is complete. Step (3): Let U = {uλ | λ ∈ Λ } be an open cover of C. Consider the set of pairs

Σ = {(n, j) | (∃λ ∈ Λ)(Vn,j ⊆ Uλ }, and notice that Σ is a countable set. Deﬁne, using the Axiom of Choice in general, ϕ :Σ → Λ by ϕ((n, j)) = µ ∈ Λ such that Vn,j ⊆ Uµ. Since Σ is countable, the set Ran(ϕ) is countable as well. Step (4): The family V = {Uλ | λ ∈ Ran(ϕ) } is a (countable) sub-cover of U. To prove this claim, we have to show that C ⊆ ∪V. To this end, let c ∈ C. There is a λ0 ∈ Λ such that c ∈ Uλ0 . By Step (2), there is a triplet (n0, j0) such that c ∈ Vn0,j0 ⊆ Uλ0 . So, (n0, j0) ∈ Σ. By the deﬁnition of ϕ we have that Vn0,j0 ⊆ Uϕ((n0,j0)), and therefore c ∈ Uϕ((n0,j0)). This immediately implies that c ∈ ∪V. Since c is any point in C, we get what we wanted to prove: C ⊆ ∪V.

The theorem is proved. 2

This theorem motivates the following deﬁnition.

Definition 7.3.2 Let C be a subset of a Hausdorff topological space (X, TX ). C is called compact if every open cover of C has a finite sub-cover. The space X is called compact if it is compact as a sub-set of itself.

We know that in R compact subsets are closed and bounded. The generalization of boundedness to a general topological space is involved, and we will skip its discussion here. On the other hand, closed subsets are given by the topology of X. We have the following result relating compact subsets to close subsets of X when X is separated.

Theorem 7.3.3 Let K be a compact sub-set of the separated topological space (X, TX ). Then, K is closed in X. Proof It is equivalent to showing that X \ K is open in X. To this end, let y ∈ X \ K. We have to show that y has a nbhd which is disjoint with K. For every x ∈ K, let Ux,Vx be open subsets such that x ∈ Ux, y ∈ Vx, and Ux ∩ Vx = ∅. Such open sets exist, because X is a separable topological space. Now, the family U = {Ux | x ∈ K } is an open cover of K, and since K is compact, U has a ﬁnite sub-cover. That is, there are x1, x2, . . . , xs ∈ K such that K ⊆ Ux1 ∪ · · · ∪ Uxs . Let

V = Vx1 ∩ · · · ∩ Vxs . Obviously y ∈ V , and

V ∩ (Ux1 ∪ Ux2 ∪ · · · ∪ Uxs ) = ∅. Therefore V ∩ K = ∅ as well, and we are done. 2

7.4 Continuity of Functions

The concept of continuity is simpler than the one of limit! Here is the ”epsilon-delta” deﬁnition of the concept. Enjoy!

Deﬁnition 7.4.1 Let f : X → R be a function, and let x0 ∈ X. We say that f is continuous at x0 if (∀ > 0)(∃δ > 0)(∀x ∈ X)((|x − x0| < δ) → (|f(x) − f(x0)| < )).

The function f is called continuous, if it is continuous at every point x0 ∈ X.

Exercise 7.4.1 (1) Prove that the function f : X → R is continuous at x0 ∈ X if, and only if, for ∗ every nbhd V of f(x0) the total pre-image P (f)(V ) is a nbhd of x0. By deﬁnition, this means that, for every such nbhd V , there is a nbhd U of x0 in R such that P∗(f)(V ) = X ∩ U.

109 (2) Prove that f : X → R is continuous if, and only if, for every open subset V ⊆ R, the total pre-image P∗(f)(V ) ⊆ X is an open subset. Again by deﬁnition this means that, for every such V , there is an open subset U ⊆ R such that P∗(f)(V ) = X ∩ U.

This exercise motivates the following deﬁnition for any topological spaces.

Deﬁnition 7.4.2 Let (X, TX ) and (Y, TY ) be two topological spaces, and let f : X → Y be a set map. (1) The map is continuous at x0 ∈ X if for every V ∈ TY such that f(x0) ∈ V , there is a U ∈ TX ∗ with x0 ∈ U and U ⊆ P (f)(V ). (2) The map is continuous if it is continuous at every point of X. Equivalently, f is continuous if ∗ for every V ∈ TY we have P (f)(V ) ∈ TX .

Returning to the case at hands, X,Y ⊆ R, notice that the only restriction on x0 in the definition of continuity of f at x0 is that x0 ∈ X. There is no need for f to be defined in a punctured nbhd of x0, as is a standard approach in popular Calculus books. Also, comparing to the defintion of limit of f at x0, we have L determined by f itself: L = f(x0), and x0 need not be an accumulation point of X. Moreover, having a limit and being continuous at x0 are independent features of f at x0 in general. For instance, it is possible for f to have a limit at x0 without being continuous at it, as well as it is possible for f to be continuous at x0 without having a limit at it! The concept of limit and of continuity of f at a point, x0, are related only if x0 ∈ X is an accumulation point of X. Here is the link between the two concepts explained.

Theorem 7.4.1 Let f : X → R, and x0 ∈ X be an accumulation point of X. Then the following statements are equivalent. (1) f is continuous at x0; 0 0 (2) If < xn > is a good sequence for f at x , then < f(xn) > converges to f(x ); 0 0 (3) f has a limit at x , and limx→x0 f(x) = f(x );

Proof. Follows directly from the deﬁnitions, and from the properties of limits of a function at a point and good sequences of a function at a point. 2

Exercise 7.4.2 Prove the previous theorem.

As a contrast verify the following.

Exercise 7.4.3 Suppose f : X → R, x0 ∈ X which is not an accumulation point of X. Then f is continuous at x0. Conclude that if X is a discrete subset of R, i.e., X has no accumulation points in R, then f is a continuous function. Moreover, f has no limit at any real number! The following theorem is a direct consequences of Theorem 7.1.3.

Theorem 7.4.2 Let f, g : X → R be two continuous at x0 ∈ X functions. Then, we have (1) f + g is continuous at x0, (2) f · g is continuous at x0, (3) If g(x0) 6= 0, then there is a δ > 0 such that (∀x ∈ X)((|x − x0| < δ) → (|g(x)| ≥ |g(x0)|/2)); (4) If g(x0) 6= 0, then f ÷ g is continuous at x0; (5) Suppose h : X → R as well, and that there is a δ > 0 such that for all x ∈ X and |x − x0| < δ, we have f(x) ≤ h(x) ≤ g(x). If f(x0) = g(x0), then h is continuous at x0 as well. Proof. Straightforward, using sequences. 2

Exercise 7.4.4 Prove the previous theorem.

110 One of most important theorems related to continuous functions is the following one. The property of continuous functions described in this theorem actually characterizes the continuous functions!

Theorem 7.4.3 (Intermediate Value Theorem) Suppose f : X → R is a continuous function, and [a; b] ⊆ X. Then

(∀y)((f(a) ≤ y ≤ f(b) ∨ f(b) ≤ y ≤ f(a)) → (∃c)(a ≤ c ≤ b ∧ y = f(c))).

Proof. If y = f(a) or y = f(b), there is nothing to be proved: we choose c = a or c = b respectively. So, let’s assume that f(a) < y < f(b). In this case, considering the function g(x) = f(x) − y, the statement of the theorem is equivalent to proving that if g(a) · g(b) < 0, then there is a c ∈ (a, b) such that g(c) = 0. We are proving therefore the following statement: Suppose g(x) is continuous in [a, b], and that g(a) · g(b) < 0. Then, there is a c ∈ (a, b) such that g(c) = 0. The idea of the proof is to define two sequences, < xn > and < yn >, of elements of (a, b) such that < xn > is increasing, < yn > is decreasing, for every n, xn < yn, the two sequences have the same limit, c, and f takes on positive values only on one of the sequences, and negative values only on the other one. All this arranged will force f(c) = 0, because if say f(xn) < 0 and f(yn) > 0 then < f(xn >→n f(c) ≤ 0, and < f(yn >→n f(c) ≥ 0. Also, notice that in this case a < xn ≤ c ≤ ym < b, so that c ∈ (a, b) as needed. Construction of the sequences < xn > and < yn >. W.L.O.G. we may assume that f(a) < 0 (and therefore that f(b) > 0). We construct the sequences by induction (recursion, actually). Define x0 = a and y0 = b. Assume we have defined x0, . . . , xn and y0, . . . , yn in such a way that b − a x ≤ · · · ≤ x , y ≥ · · · ≥ y , y − x = , f(x ) < 0 < f(y ) k = 0, . . . n. 0 n 0 n k k 2k k k 0 We are defining xn+1 and yn+1 as follows. Consider the midpoint x = (xn + yn)/2 of the interval 0 0 0 0 [xn, yn]. If f(x ) = 0, then we choose c = x , and are done. So, assume that f(x ) 6= 0. If f(x ) < 0, 0 0 0 we define xn+1 = x and yn+1 = yn. If (x ) > 0, we define xn+1 = xn, and yn+1 = x . In any case, we have that b − a x ≤ x , y ≥ y , f(x ) < 0 < f(y ), y − x = n n+1 n n+1 n+1 n+1 n+1 n+1 2n+1 as needed. This way we construct the two sequences, and therefore complete the proof. 2

Exercise 7.4.5 (∗) The important Intermediate Value Theorem can be proved many ways. One of them is sketched here. (1) W.L.O.G., we may assume that f(a) < y < f(b). Consider in this case the set

Σ = {r | (∀x)(x ∈ (a, r) → f(x) < y)} and prove that it is non-empty, and bounded above. 0 00 (2) Let s be the least upper bound of Σ. Construct good sequences for f at s, < xn > and < xn >, 0 00 such that f(xn) < y < f(xn). (3) Conclude the proof using that f is continuous at s, and applying item (5) of Theorem 6.3.1. This theorem has a geometric (topological) meaning.

Corollary 7.4.4 Suppose X ⊆ R is an interval, no matter open, closed or semi-open/semi-closed. If f : X → R is a continuous function, then Ran(f) ⊆ R is also an interval. That is, if y1, y2 ∈ Ran(f) with y1 < y2, then (∀y)((y1 < y < y2) → (∃x ∈ X)(y = f(x))).

Proof. To a, b ∈ X such that y1 = f(a) and y2 = f(b) apply the Intermediate Value Theorem. 2

Exercise 7.4.6 Complete the proof of the previous corollary.

111 As mentioned before, the compact subsets of R are very useful when working with continuous functions. This is so, partially for the following reason.

Theorem 7.4.5 Let X ⊆ R be a subset, K ⊆ X be a compact subset, and let f : X → R be a continuous function. Then f(K) ⊆ R is also compact.

Proof. We need to show that every sequence < yn >⊆ f(K) has a convergent in f(K) subsequence.. But there is a sequence < xn >⊆ K such that (∀n)(yn = f(xn)). Since K is compact, there is a 0 0 subsequence < xnk >→k x ∈ K. Since f is continuous, then < f(xnk ) >→k f(x ) ∈ f(K). So,

< ynk > is a subsequence of < yn > which is convergent in f(K) as needed. 2

Exercise 7.4.7 Prove this theorem using the equivalent description of compact subsets in terms of open covers. Here is a theorem which provides an excellent ﬁnale to our Introduction to Advanced Calculus part of the Notes!

Theorem 7.4.6 (Weierstrass) Let f : X → R be a continuous function with X ⊆ R - a compact subset. Then, f attains its minimum and maximum values on X. In other words,

(∃x1, x2 ∈ X)(∀x ∈ X)(f(x1) ≤ f(x) ≤ f(x2)).

Proof. We know that Ran(f) is compact in R, so it is closed and bounded. The boundedness gives us that Ran(f) has l.u.b., M, and g.l.b., m. The closeness of Ran(f) gives us that m, M ∈ Ran(f). Note that m is the minimum, and that M is the maximum of f on X. 2

Example 7.4.1 Since [a, b] ⊆ R is closed and bounded, then it is compact. If f :[a, b] → R is a continuous function, then f attains its extreme values, the minimum and the maximum, on [a, b]. 2

Exercise 7.4.8 Let f :[a, b] → R be a continuous function. Prove that Ran(f) = [m, M] where m is the minimum, and M is the maximum of f on [a, b].

Q.E.D.

112 Index

< xn >→n L, 93 Chain, R-chain, 63 S(X), 26 Choice function for a set X, 54 T1-space, 92 Classical topology on R, deﬁnition, 91 T2-space, 91 Classical topology on R, epsilon nbhd of a, 90 X ∩ Y , 25 Classical topology on R, open subset, 90 X ⊕ Y , 27 co-domain of a (set) relation, co − Dom(R), X \ Y , 26 45 X ⊆ Y , 22 Commutative group, 68 N, 26 Commutative ring, 68 Ω, the class of all sets, 21 Comparable elements, (R-), 63 R, 82 Composable (set) functions, 51 R−, R+, 84 Composition of relations, S ◦ R, 47 RC, 103 Conclusion/consequent, 5 Z+, Z≥0, Z−, 73 Conditional, →, 5 ∩X, 25 Conjunction, ∧, 5 ∅, 22 Connectives, Logic, 5, 6 − δ. limit, 104 Connectives, Math, 10 F(X,Y ),X,Y ⊆ R, 104 Contradiction, Logic, 7 P∗(f), 52 Contradiction, method of proof by, 17 P∗(f), 52 Contrapositive method of proof, Logic, 16 {x, y}, 24 Contrapositive, Logic, 5 f −1, 53 Counterexample to a universal f −1(Y ), 52 quantiﬁcation, 11 iA, 46 De Morgan laws, Sets, 27 Accumulation point of A, 98 Dedekind cut, 82 Actual variables, Logic, 10 Dedekind cut, irrational, 83 Algebraic numbers, A, 80 Dedekind cut, rational, 82 Alphabet, Logic, 4, 6 Delimiters, Logic, 5, 6 AM-GM inequality, 41 Denying propositional functions, 10 Apparent variables, Logic, 10 Direct method of proof, Logic, 15 Archimedean property of R, 86 Discrete topology on X, 91 Arithmetic operations on N, 36 disjunction, ∨, 5 Axiom of Choice, 54 Disjunctive syllogism, rule of inference, 14 Axiom of foundation, 65 Doubleton, {x, y}, 24

Biconditional proposition, Logic, 5 Empty set axiom, Set Theory, 22 Bolzano-Weierstrass theorem, for multisets, Empty set, ∅, 22 99 Essentially unique, 68 Bolzano-Weierstrass theorem, for sets, 96 Exclusive or, XOR, 6 Bound variables, Logic, 10 Existential generalization, rule of inference, Bounded above, 62 15 Bounded below, 62 Existential instantiation, rule of inference, 15

113 Existential quantification, 11 Modus tollens, denies by denying, rule of Existential quantifier, 11 inference, 14 Extensionality axiom, Set Theory, 22 Monotone homomorphism, 77 Monotone map, 68 Finite Complement Topology on X, 92 Multi-sets, 55 Fréchet space, 92 Multiset of real numbers, accumulation point Free variables, Logic, 10 of, 98 Function, choice, 54 Function, correspondence, 47 Natural numbers, N, 26 Function, inverse, f −1, 53 Natural numbers, definition, 38 Function, left inverse, 54 Negation/denial, 6=, 5 Function, multi-valued, 47 Function, one-to-one, 49 Open sentences, 10 Function, one-to-one Operations, binary, Math, 10 correspondence/bijection, 53 Ordered field, complete, 81 Function, one-to-one/injection, 53 Ordered field, completion of, 81 Function, onto, 49 Ordered pair, (x, y), 43 Function, onto/surjection, 53 Function, right inverse, 54 Pairing axiom, Set Theory, 23 Partition of a set, 57 Greatest element, 61 Peano’s axioms, 39 Greatest lower bound, g.l.b., 62 PMI, 33 Power set axiom, Set Theory, 23 Hausdorff space, 91 Power set of X, P(X), 23 Homomorphism, 68 Predecessor, immediate, N, 33 hypothesis/condition/antecedent, 5 Product of two sets (Cartesian), X × Y , 45 Hypothetical syllogism, rule of inference, 14 proposition, 4 Propositional expressions, 6 Identity relation, iA, 46 Propositional functions, Logic, 9 iff, 5 Image function associated with f, P∗(f), 52 Quantifiers, Logic, 9, 11 Index set, 55 Quotient map, πR : A → A/R, 58 Indexed family of sets, 55 Quotient set, A/R, 58 Inductive set, 27 Infinity axiom, Set Theory, 27 RAA, 17 Initial segment, 62 Rational numbers, a construction, 75 Integers, a construction, 69 Rational numbers, definition, 74 Integers, definition, 68 Real numbers, a construction, Cantor’s Intermediate Value Theorem, 111 approach, RC, 102 Intersection of X, ∩X, 25 Real numbers, a construction, Dedekind’s Inverse relation, (B, A, R−1), 46 approach, R, 82 Real numbers, definition, 81 Least element, 61 Real valued functions of a real variable, Least element principal, LEP, on N, 32 continuity of, at a point, 109 Least upper bound, l.u.b., 62 Real valued functions of a real variable, LEP, 32 Intermediate Value Theorem, 111 Logically equivalent propositions, Logic, 6 Real valued functions of a real variable, limit Long division for integers, 40 of, at a point, − δ, 104 Lower bound, 62 Real valued functions of a real variable, limit of, sequential approach, 105 Maximal element, 61 Real valued functions of a real variable, Metrics space, (X, d), 90 Squeeze Theorem, 106 Minimal element, 61 Real valued functions of a real variable, Modus ponens, law of detachment, rule of Weierstrass’s theorem, 112 inference, 14 Recursion, Set theory, 36

114 Reductio Ad Absudrdum, RAA, 17 Tautology, Logic, 7 Relation from A to B, 45 Toplogy on a set, discrete topology on X, 91 Relation, anti-symmetric, 56 Topology non a set, accumulation point, 95 Relation, asymmetric, 56 Topology on a set, TX , 91 Relation, connected, 56 Topology on a set, closed subset, 92 Relation, continuous order, 61 Topology on a set, compact subsets, 109 Relation, dense order, 61 Topology on a set, compact subsets of R, 107 Relation, equivalence, 57 Topology on a set, continuous map, at a Relation, greatest lower bound property of, point, 110 61 Topology on a set, finite complement Relation, irreflexive, 56 topology on X, 92 Relation, least upper bound property of, 61 Topology on a set, open covers/subcovers of Relation, partial order, 61 a set, 108 Relation, partial order, strict, 61 Topology on a set, open subset, 91 Relation, pre-order, 60 Topology on a set, topological space, Relation, reflexive, 56 (X, TX ), 91 Relation, symmetric, 56 Topology on a set, trivial topology on X, 91 Relation, total order, 61 Topology on a set, Zariski open subsets of R, Relation, total order, strict, 61 91 Relation, transitive, 56 Topology on a set, Zariski toplogy on R, 91 Relation, trichotomous, 56 Total preimage function associated with f, Relation, well order, 61 P∗(f), 52 Relations, binary, Math, 10 Total preimage of a set, f −1(Y ), 52 Relative complement of Y in X, X \ Y , 26 Transfinite induction, 35 Resolution, rule of inference, 14 Transitive set, 26 Restriction of a (set) relation, R,C×D46 Trichotomy property, 33 Reversed/backward induction, 35 Trivial topology on X, 91 Russell’s class axiom, Set Theory, 25 Truth set of a propositional function, 11 Truth table, Logic, 8 Separated space, 91 Separation Axiom, Set Theory, 25 Union axiom, Set Theory, 24 Sequence in , good sequence for f at x0, 105 R Union of X and Y , X ∪ Y , 24 Sequence in , subsequence of, 98 R Union of X, ∪ X, 24 Sequence in a set, 93 Universal generalization, rule of inference, 15 Sequence in a set, convergent/converges to, 93 Universal instantiation, rule of inference, 15 Sequence in a set, limit of, 93 Universal quantification, 11 Sequence in a set, stationary/constant, 93 Universal quantifier, 11 Sequence of real numbers, Cauchy, 95 Universe of all propositions, 9 Sequence of sets, finite, infinite, 55 Upper bound, u.b., 61 Sequences of real numbers, (strictly) increasing/decreasing/monotone, 94 Variables, Logic, 6 Sets, 21 Singleton{x}, 24 Weierstrass’s Theorem, 112 Squeeze Theorem, 106 Witnesses/meanings of a variable, 11 Subset, 22 Subset, proper, 22 Zermelo’s theorem, 63 Successor of a set, S(X), 26 Zermelo-Fraenkel axiomatic Set Theory, ZF, Symmetric difference of the sets X and Y , 65 X ⊕ Y , 27 ZF, 65 ZFC, 65 Tautologies, 6 Zorn’s lemma, 63

115